It suffices to prove this when the $A$ is mapped to a stochastic matrix, i.e. divide out the Perron root, and do the diagonal similarity transform here
$\rho(B) \leq \rho(|B|)$, where $\rho$ is the spectral radius
The result is a matrix associated with an irreducible Markov chain. Call this stochastic matrix $P$.
Note: when I say a Markov matrix can be written as a direct sum, technically what is meant is that up to graph isomorphism (i.e. permutation similarity) it is direct sum a / direct product / block diagonal. $P^k$ being reducible for some $k$ alway implies block diagonal (as opposed to mere block triangular) since $P^k$ cannot have transient states since $P$ is irreducible (check e.g. $\pi P = \pi=\pi P^k$ and $\pi$ is a positive vector)
OP has already stated that the aperiodic case is understood--so that is taken for granted. Note: aperiodicity is necessarily equivalent to there being only one eigenvalue on the unit circle. Thus this proof starts only with the coarse association that $\geq 2$ eigenvalues on the unit circle for irreducible $P$ is equivalent to $P$ having some period $\geq 2$.
Let $P$ a stochastic matrix with index of imprimitivity $=r\geq 2$. This means there are $r$ eigenvalues on the unit circle, only one of which $=1$ since $A$ is irreducible. The boundedness of $P$ (using e.g. Frobenius norm or operator 2 norm) implies these eigenvalues are all necessarily semi-simple (why?) and we'll prove they are in fact simple, forming a complete set of $r$th roots of unity. For expediency, we write out the Jordan Form
$P = S\pmatrix{D_r & \mathbf 0\\\mathbf 0&J}S^{-1}$
where $D_r$ is a $r\times r$ diagonal matrix containing the eigenvalues on the unit circle and $J$ contains Jordan blocks associated with eigenvalues strictly inside the unit circle. We now put this aside for a bit and prove some lemmas.
Lemma 1:
Let $P$ be an irreducible $n\times n$ stochastic matrix with period $\geq 2$. Then $\text{trace}\big(P\big)=0$.
(The $n=1$ case trivially has trace one and is trivially aperiodic hence excluded). For $n\geq 2$, by stochasticity we know $\text{trace}\big(P\big)\geq 0$, so suppose for contradiction that $\text{trace}\big(P\big)\gt 0$. Then $P$ has exactly one $\lambda = 1$ (Perron root) and at least one other eigenvalue $\omega\neq 1$ which is on the unit circle (if no $\omega$ exists, then $\lim_{k\to\infty}P^k$ exists and $P$ is aperiodic). Further $\text{trace}\big(P\big)\gt 0\implies$ at least one diagonal of $P$ that is positive, i.e. some $p_{j,j}\gt 0$. Then by applying (reverse) triangle inequality to row $j$ we see $\big(\omega I - P\big)$ is weakly diagonally dominant and irreducible, hence by Taussky's refinement of Gerschgorin Discs $\det\big(\omega I - P\big)\neq 0$ and $\omega$ is not an eigenvalue, a contradiction. This proves the lemma.
(I gave a proof of Taussky's refinement under "optional second", here: Prove that this block matrix is positive definite .)
Lemma 2:
all eigenvalues on the unit circle for irreducible $P$ are roots of unity (i.e. each satisfies $\lambda^k=1$ for some positive integer $k$).
If $P$ is aperiodic then the result is immediate. Suppose $P$ is periodic. We do this via strong induction on $n$. The (Base) $2\times 2$ case: $P$ is stochastic and trace zero hence $P= \pmatrix{0 & 1\\1&0}$
Inductive case:
We know $\text{trace}\big(P\big) =0$ but $\text{trace}\big(P^k\big) \gt 0$ for some minimal $k\in \big\{2,3,...,n\big\}$. Note: $\text{trace}\big(P^k\big) \lt 0$ is impossible, and $\text{trace}\big(P^k\big) = 0$ for all $1\leq k \leq n$ powers implies $P$ is nilpotent which contradicts irreducibility. Thus using the above minimal $k$, via Lemma 1 we know $P^k$ is reducible and/or aperiodic. The former case explicitly says $P^k$ is reducible and we show it must be in the latter case as well. If $P^k$ is aperiodic, all eigenvalues on the unit circle $=1$ which implies $\omega^k=1$ for some $\omega \neq 1$, and by Perron-Frobenius Theory $P^k$ is thus reducible into a collection of irreducible stochastic matrices, each of which has a single eigenvalue on the unit circle if aperiodic, and for period $\geq 2$ we may apply our induction hypothesis. Thus each eigenvalue on the unit circle is a $k_i$th root of unity so $k':=k_1\cdot k_2\cdot ... \cdot k_r$ and they all satisfy $x^{k'}-1=0$.
Lemma 3:
Now consider the projection: $T = S\pmatrix{I_r & \mathbf 0\\\mathbf 0&\mathbf 0}S^{-1}$. Then
$M:=TP=PT= S\pmatrix{D_r & \mathbf 0\\ \mathbf 0&\mathbf 0}S^{-1}$
is a stochastic matrix.
Note that $T= \lim_{s\to \infty} \big(P^{k'}\big)^s$
which is a stochastic matrix and the product of two stochastic matrices is a stochastic matrix
Lemma 4:
$0 \leq \text{trace}\big(M^k\big) = \text{trace}\big(D_r^k\big)\leq r$
$M$ is a stochastic matrix hence $0\leq \text{trace}\big(M^k\big)$ and $\text{trace}\big(M^k\big) = \text{trace}\big(D_r^k\big)$ since $M$ and $D_r$ have the same non-zero eigenvalues.
By triangle inequality
$\text{trace}\big(M^k\big) =\text{trace}\big(D_r^k\big) = \big \vert \text{trace}\big(D_r^k\big) \big \vert \leq \sum_{k=1}^r \vert d_{i,i}\vert^k = r$ with equality iff $D_r^k = I$.
Lemma 5 (using character theory):
(i) $\text{trace}\big(M^k\big) \in \big\{0,r\big\}$ for all $k$ and (ii) all non-zero eigenvalues of $M$ are distinct.
Let $G$ be the cyclic group generated by $D_r$. Since this is already a matrix group then we may call $G'$ a representation of $G$ via what amounts to the identity homomorphism. I.e. $G'$ is a (faithful) $r$-dimensional representation of a finite abelian matrix group, $G$. Being abelian all irreducible representations are $1$-dimensional and the number of irreducible representations $=\vert G \vert$. The character associated with $G'$, $\chi$, may be written as
$\chi = \eta_1 \cdot\chi_1+\eta_2\cdot \chi_2 + ... + \eta_{\vert G \vert}\cdot \chi_{\vert G \vert}= 1\cdot \chi_1+\eta_2 \cdot\chi_2 + ... + \eta_{\vert G \vert} \cdot\chi_{\vert G \vert}$
where $\eta_j \in \mathbb N$ (inclusive of zero). (In terms of notation, page 319 of Artin's Algebra 1st edition could be useful here.)
We know the multiplicity of eigenvalue 1 for $G'$ (the trivial representation) is $1$. Using the standard inner product for characters we have
$1=\langle \chi_1, \chi \rangle= \frac{1}{\vert G \vert}\sum_{k=1}^{\vert G\vert}1 \cdot \text{trace}\big(D_r^k\big)$
re-scaling by $r$ we have
$r= \frac{1}{\vert G \vert}\sum_{k=1}^{\vert G\vert}r \cdot \text{trace}\big(D_r^k\big)$
$\geq \frac{1}{\vert G \vert}\sum_{k=1}^{\vert G\vert} \text{trace}\big(D_r^k\big)\cdot \text{trace}\big(D_r^k\big)=\langle \chi, \chi\rangle = \eta_1^2+\eta_2^2 + ... + \eta_{\vert G \vert}^2$
$\geq \sum_{j=1}^{\vert G\vert} \eta_j=r$
The first inequality comes from the prior lemma and is met with equality iff $\text{trace}\big(D_r^k\big) \in \big\{0,r\big\}$ for all $k$ and the second inequality comes from the fact that $\eta_j \in \mathbb N$ and is met with equality iff all $\eta_j \in \big\{0,1\big\}$ i.e. if all eigenvalues of $D_r$ are distinct.
Since both inequalities are met with equality we conclude that all eigenvalues are distinct and $\text{trace}\big(M^k\big) \in \big\{0,r\big\}$ for all $k$
Lemma 6:
$D_r$ consists of a complete set of $r$th roots of unity.
Since $M$ is similar to $\pmatrix{D_r & \mathbf 0\\ \mathbf 0&\mathbf 0}$ and all eigenvalues of $D_r$ are distinct, we know $M$ has minimal polynomial
$M^{r+1} + c_r M^r + c_{r-1}M^{r-1}+ ...+c_1 M=\mathbf 0$
i.e. $M^{r+1}$ is written uniquely as linear combination of $\big\{M^r, M^{r-1}, ..., M\big\}$. Put differently, $\big\{M^r, M^{r-1}, ..., M\big\}$ provides a basis for powers of $M$. Since $\text{trace}\big(M^k\big)\gt 0$ for some $k$ ($M$ is not nilpotent), the above basis in combination with the Lemma $5$ implies
$\text{trace}\big(M^k\big)=r$ for some $k\in \big\{1,2,...,r\big\}$
Lemma $4$ tells us that $\text{trace}\big(M^k\big)=r\implies D_r^k = I_r \implies$ all eigenvalues of $D_r$ are $k$th roots of unity. However for $k \in \big\{1,2,...,r-1\big\}$ there are $\lt r$ distinct kth roots of unity and all eigenvalues of $D_r$ are distinct by the lemma 5 $\implies \text{trace}\big(M^k\big)=0 $ for $k \in \big\{1,2,...,r-1\big\}\implies \text{trace}\big(M^r\big)=r\implies D_r$ consists of a complete set of $r$th roots of unity.
Corollary:
$\text{trace}\big(M^k\big) = 0$ if $k\% r \neq 0$ (i.e. when $k$ is not a multiple of $r$)
$\text{trace}\big(M^k\big) = r$ when $k\% r = 0$ (i.e. when $k$ is a multiple of $r$)
main problem:
Suppose some diagonal element of $P^k$ is positive, then $\text{trace}\big(P^k\big)\gt 0$, and $P^k$ is reducible by the argument in Lemma 2 (in the induction case). So $P^k$ is a direct sum of $m=m_0+m_1$ irreducibles, where $m_0$ of those have period $p_i\geq 2$ each having trace zero by lemma 1, thus $\text{trace}\big(P^k\big)= $ sum of traces of $m_1$ aperiodic irreducibles (each having a single eigenvalue with modulus 1, i.e. the Perron root).
define $T':= \lim_{t\to \infty}\Big(\big(P^k\big)^{k^{''}}\Big)^t$
where mimicking before, $k^{''}$ is chosen to send all unit modulus eigenvalues of $\big(P^k\big)$ to 1.
When we apply $T'$, we get $\big(T'P^k\big)$ is still a direct sum of $m = m_0+m_1$ irreducibles
i.e. taking a power of $\big(P^k\big)$ respects the block diagonal structure it has (up to graph isomorphism) and this necessarily holds in limits as well. When we examine an arbitrary block of $\big(P^k\big)$ we see that $T'$ kills all subdominant spectra but acts as the identity on roots of unity hence periodicity ($m_0$) and aperiodicity ($m_1$) is preserved.
but $T'=S\pmatrix{I_r & \mathbf 0\\ \mathbf 0&\mathbf 0}S^{-1}=T$ thus
$T'P^k= TP^k = \big(TP\big)^k=M^k$ so
$ \text{trace}\big(M^k\big)=m_1\gt 0\implies k$ is a multiple of $r$ by the preceding corollary. Finally $J^k\to \mathbf 0$ means $\big\vert \text{trace}\big(M^{r\cdot t}\big)- \text{trace}\big(P^{r\cdot t}\big)\big\vert\lt \epsilon$ for any $\epsilon \gt 0$ for $t$ large enough, in particular after selecting $t$ we have $\text{trace}\big(P^{r\cdot t}\big)\gt 0$ and $\text{trace}\big(P^{r\cdot {(t+1)}}\big)\gt 0$ thus $r$ is the GCD. We have established a bijection that $r$ eigenvalues on the unit circle for $P$ implies period $r$ for arbitrary integer $r\geq 2$ and this completes the proof.
