I've just been able to prove that my expectation is true.
In order to prove this, let us define the following sequence for $n\in\mathbb N$ :
$$e_{1,n}=n+1,\ \ e_{m,n}=ne_{1,n}e_{2,n}\cdots e_{m-1,n}+1\ \ (m=2,3,4,\cdots).$$
Lemma 1 :
$$\begin{align}\frac1{e_{1,n}}+\frac1{e_{2,n}}+\cdots+\frac1{e_{m,n}}=\frac1n-\frac1{e_{m+1,n}-1}.\end{align}$$
Proof for lemma 1 : By the definition, we get
$$(e_{m-1,n}-1)e_{m-1,n}=ne_{1,n}\cdots e_{m-2,n}\cdot e_{m-1,n}=e_{m,n}-1.$$
Since
$$\frac1{e_{m-1,n}}=\frac1{e_{m-1,n}-1}-\frac1{e_{m,n}-1},$$
we get
$$\sum_{i=2}^{m+1}\frac1{e_{i-1,n}}=\sum_{i=2}^{m+1}\left(\frac1{e_{i-1,n}-1}-\frac1{e_{i,n}-1}\right)=\frac1{e_{1,n}-1}-\frac1{e_{m+1,n}-1}=\frac1n-\frac1{e_{m+1,n}-1}.$$
Now the proof for lemma 1 is completed.
Lemma 2 :
$$ x_1y_1+x_2y_2+\cdots+x_my_m $$$$=(x_1-x_2)y_1+(x_2-x_3)(y_1+y_2)+\cdots+(x_{m-1}-x_m)(y_1+y_2+\cdots+y_{m-1})+x_m(y_1+y_2+\cdots+y_m)$$
Proof for lemma 2 : Let $S_i=y_1+y_2+\cdots+y_i\ (i\in\mathbb N)$.
Then, we get
$$\begin{align}\sum_{i=1}^mx_iy_i & = x_1y_1+\sum_{i=2}^mx_i(S_i-S_{i-1}) \\
& = x_1y_1+\sum_{i=2}^mx_iS_i-\sum_{i=1}^{m-1}x_{i+1}S_i \\
& = x_1y_1+\sum_{i=2}^{m-1}(x_i-x_{i+1})S_i+x_mS_m-x_2S_1 \\
& = (x_1-x_2)y_1+\sum_{i=2}^{m-1}(x_i-x_{i+1})(y_1+\cdots+y_i)+x_m(y_1+\cdots+y_m).
\end{align}$$
Now the proof for lemma 2 is completed.
Lemma 3 : If
$$\frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_m}\lt\frac1n$$
where $x_i\ (i=1,2,\cdots,m)$ are natural numbers, then
$$\begin{align}\frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_m}\le\frac1{e_{1,n}}+\frac1{e_{2,n}}+\cdots+\frac1{e_{m,n}}\qquad(\star)\end{align}$$
Proof for lemma 3 : Let us prove this by induction on $m$. We'll use the proof by contradiction. Also, by the symmetry about the letters, we may suppose that $x_1\le x_2\le \cdots\le x_m.$
If $m=1$, then since $x_1\gt n$ and $x_1\ge n+1=e_{1,n}$, we get $(\star).$
In the following, let's suppose $(\star)$ for any less than or equal to $m-1$.
Now, for $m$, let's suppose that there exists $(x_1,x_2,\cdots,x_m)$ which does not satisfy $(\star)$. (we'll use the proof by contradiction)
This means $(x_1,x_2,\cdots,x_m)$ satisfies
$$\begin{align}\frac1{e_{1,n}}+\frac1{e_{2,n}}+\cdots+\frac1{e_{m,n}}\lt \frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_m}\lt \frac1n\qquad(\star\star)\end{align}$$
Hence, from the lemma 1, we know
$$0\lt\frac1n-\left(\frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_m}\right)\lt\frac1n-\left(\frac1{e_{1,n}}+\frac1{e_{2,n}}+\cdots+\frac1{e_{m,n}}\right)=\frac1{e_{m+1,n}-1}.$$
Here, we know, by reducing the fractions to a common denominator,
$$\frac1n-\left(\frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_m}\right)\ (\gt 0)$$
can be represented as
$$\frac{M}{nx_1x_2\cdots x_m}$$
where $M\in\mathbb N.$
By the definition of the sequence $\{e_{m,n}\}$, we get
$$\frac{1}{nx_1x_2\cdots x_m}\le\frac{M}{nx_1x_2\cdots x_m}\lt\frac{1}{ne_{1,n}e_{2,n}\cdots e_{m,n}}.$$
Hence, we get
$$\begin{align}x_1x_2\cdots x_m\gt e_{1,n}e_{2,n}\cdots e_{m,n}\qquad (1)\end{align}$$
Here, letting
$$P=\frac{x_1}{e_{1,n}}+\frac{x_2}{e_{2,n}}+\cdots+\frac{x_m}{e_{m,n}},$$
we get the following by AM–GM inequality and $(1)$ :
$$\begin{align} P\ge m\sqrt[m]{\frac{x_1x_2\cdots x_m}{e_{1,n}e_{2,n}\cdots e_{m,n}}}\gt m\qquad (2)\end{align}$$
On the other hand, we get the following by lemma 2 :
$$\begin{align} P= & (x_1-x_2)\frac1{e_{1,n}}+(x_2-x_3)\left(\frac1{e_{1,n}}+\frac1{e_{2,n}}\right)+\cdots \\ & +(x_{m-1}-x_m)\left(\frac1{e_{1,n}}+\frac1{e_{2,n}}+\cdots+\frac{1}{e_{m-1,n}}\right) \\ & +x_{m}\left(\frac1{e_{1,n}}+\frac1{e_{2,n}}+\cdots+\frac1{e_{m,n}}\right) \end{align}$$
Here, noting that $x_i-x_{i+1}\le0\ (i=1,2,\cdots,m-1)$ and that the inductive assumption leads
$$\frac1{x_1}\le\frac{1}{e_{1,n}},$$
$$\frac1{x_1}+\frac{1}{x_2}\le\frac{1}{e_{1,n}}+\frac1{e_{2,n}},$$
$$\vdots$$
$$\frac1{x_1}+\cdots+\frac{1}{x_{m-1}}\le\frac1{e_{1,n}}+\cdots+\frac1{e_{m-1,n}},$$
we get
$$\begin{align} P\le & (x_1-x_2)\frac1{x_1}+(x_2-x_3)\left(\frac1{x_1}+\frac1{x_2}\right)+\cdots \\ & +(x_{m-1}-x_{m})\left(\frac1{x_1}+\cdots+\frac1{x_{m-1}}\right) \\ & +x_m\left(\frac1{e_{1,n}}+\frac1{e_{2,n}}+\cdots+\frac1{e_{m,n}}\right).\end{align}$$
Here, using $(\star\star)$, since we get
$$\begin{align}P\lt & (x_1-x_2)\frac1{x_1}+(x_2-x_3)\left(\frac1{x_1}+\frac1{x_2}\right)+\cdots \\ & +(x_{m-1}-x_{m})\left(\frac1{x_1}+\cdots+\frac1{x_{m-1}}\right)\\ & +x_m\left(\frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_m}\right),\end{align}$$
we get $P\lt 1+1+\cdots+1=m$, which contradicts $(2)$. Hence, by the proof of contradiction, we now know that there does not exist $(x_1,x_2,\cdots,x_m)$ which satisfies $(\star\star)$. Hence, the above leads $(\star)$ for $m$. The equality is attained when $x_1=e_{1,n}, x_2=e_{2,n},\cdots, x_m=e_{m,n}.$ Now the proof for lemma 3 is completed.
Proof for my expectation : Let us prove that
$$\begin{align}\max_{\mathbf x\in S_{m,n}}\left(\max_{1\le i\le m}x_i\right)=e_{m,n}-1\qquad (3)\end{align}$$
If we can prove $(3)$, then $k_{m,n}=e_{m,n}-1$ leads
$$k_{m,n}=(k_{m-1,n}+1)k_{m-1,n}$$
by the relational expression about $\{e_{m,n}\}.$
Let us prove $(3)$ by induction on $m$.
If $m=1$, then we leads $x_1=n=e_{1,n}-1.$
In the $m\ge 2$ case, we may suppose that $x_1\le x_2\le \cdots\le x_m$.
Since
$$\sum_{i=1}^{m-1}\frac1{x_i}\lt\frac1n,$$
we get by lemma 3
$$\frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_{m-1}}\le\frac1{e_{1,n}}+\frac1{e_{2,n}}+\cdots+\frac1{e_{m-1,n}}.$$
Hence, we get by lemma 1
$$\frac1{x_m}=\frac1n-\sum_{i=1}^{m-1}\frac1{x_i}\ge\frac1n-\sum_{i=1}^{m-1}\frac1{e_{i,n}}=\frac1{e_{m,n}-1}.$$
Since $x_m\le e_{m,n}-1$, we get
$$\max_{\mathbf x\in S_{m,n}}\left(\max_{1\le i\le m}x_i\right)\le e_{m,n}-1.$$
By the way, since we know by lemma 1
$$(e_{1,n},e_{2,n},\cdots,e_{m-1,n},e_{m,n}-1)\in S_{m,0},$$
the equality of the above inequality is obtained. Now the proof for $(3)$ is completed. Hence, we now know that the proof for my expectation is completed.
