Perhaps surprisingly, the answer is no!
First, we prove the following weaker result:
Lemma: the only complete $\omega$-consistent theory containing "enough arithmetic" (Robinson's arithmetic $\mathsf{R}$ is enough) is $T=Th(\mathbb{N})$.
Proof.$\quad$Let $T$ be a complete $\omega$-consistent theory containing $\mathsf{R}$. We'll prove by induction on $k$ that for each $k$, $T$ proves every true $\Pi_k$ sentence and every $\Sigma_k$ sentence.
The base case $k=1$ follows by $\Sigma_1$-completeness of $\mathsf{R}$: our theory $T$ proves every true $\Sigma_1$ sentence, and therefore in order to be complete and $\omega$-consistent it must also prove every $\Pi_1$ sentence.
Next, we suppose the result holds for $k$ and show that it also holds for $k+1$:
Fix a $\Sigma_k$-formula $\psi(x)$ and suppose the $\Pi_{k+1}$ sentence $\forall x\psi(x)$ is true; we want to show that $T\vdash \forall x\psi(x)$. Suppose otherwise. Then by completeness, $T\vdash\exists x\neg\psi(x)$. But since $\forall x\psi(x)$ is true, by the induction hypothesis we have $T\vdash\psi(\underline{n})$ for each $n$. And this contradicts $\omega$-consistency.
Fix a $\Pi_k$-formula $\psi(x)$ and suppose the $\Sigma_{k+1}$ sentence $\exists x\psi(x)$ is true; we want to show that $T\vdash\exists x\psi(x)$. Since $\exists x\psi(x)$ is true, let $n$ be such that $\psi(\underline{n})$ is true. The sentence $\psi(\underline{n})$ is $\Pi_k$ and true, hence by induction we have $T\vdash\psi(\underline{n})$ and so $T\vdash\exists x\psi(x)$ by existential generalization.
This finishes the proof. $\quad\Box$
As an aside, note that only the first bulletpoint used $\omega$-consistency. More generally, if $S$ is a theory proving all true $\Pi_k$ sentences, then $S$ also proves all true $\Sigma_{k+1}$ sentences. This is an interesting asymmetry in the arithmetic hierarchy between the "$\Pi$" and "$\Sigma$" sides.
We now argue as follows:
Corollary: there is an $\omega$-consistent theory containing "enough arithmetic" which has no $\omega$-consistent completion.
Proof.$\quad$ Suppose otherwise. Then we have that for every sentence $\varphi$, exactly one of the following holds:
Clearly we can't have them both hold; conversely, if they both failed we would get a complete $\omega$-consistent theory containing a false sentence, contradicting the lemma.
But this means that $Th(\mathbb{N})$ can be computed from $\{\varphi:\neg\omega Con(\mathsf{R}+\varphi)\}$. But this can't be: the latter is $\Pi_3$-definable while the former isn't even arithmetic.
