Prove that $g(x)=\frac{\ln(S_n (x))}{\ln(S_{n-1}(x))}$ is increasing in $x$, where $S_{n}(x)=\sum_{m=0}^{n}\frac{x^m}{m!}$

Note: Here is a first step towards a complete answer. The point I like to address is the discrete log-concavity of the function $S_n$ for fixed $x>0$. This means $S_n$ fulfils

\begin{align*} S_{n-1}^2(x)\geq S_n(x)S_{n-2}(x)\qquad\qquad x>0, n\geq 2\tag{1} \end{align*}

A corresponding example can be found e.g. in the slides about Log-convexity and Log-concavity by Dmitry Karp. You might have a look at the section about Bessel functions where the inequality (1) for these functions is stated and some sharper results are presented afterwards.

Please note that in order to show $g(x)$ is increasing an even sharper inequality than (1) has to be proved.

We start as @Arashium did. Using the differential operator $D_x$ we obtain \begin{align*} D_xS_n(x)=D_x\sum_{m=1}^{n}\frac{x^{m-1}}{(m-1)!}=\sum_{m=0}^{n-1}\frac{x^m}{m!}=S_{n-1}(x) \end{align*} and by omitting the argument $x$ we get \begin{align*} D_x\ln(S_n)&=\frac{1}{S_n}D_x(S_n)=\frac{S_{n-1}}{S_n}\\ \end{align*} and \begin{align*} D_xg&=D_x\frac{\ln(S_n)}{\ln(S_{n-1})} =\frac{\ln(S_{n-1})D_x\ln(S_n)-\ln(S_n)D_x\ln(S_{n-1})}{\ln^2(S_{n-1})}\\ &=\frac{1}{\ln^2(S_{n-1})}\left(\frac{S_{n-1}}{S_n}\ln(S_{n-1})-\frac{S_{n-2}}{S_{n-1}}\ln(S_n)\right)\tag{2} \end{align*}

In order to show that the function $g$ is increasing we need $D_xg\geq 0$.

Therefore the following is to prove according to (2) \begin{align*} S_{n-1}^2\geq S_{n-2}S_n\frac{\ln(S_n)}{\ln(S_{n-1})}\tag{3} \end{align*}

Here's the proof of the weaker inequality (1) showing that $S_n(x)$ is discrete log-concave for fixed $x>0$.

\begin{align*} S_{n-1}(x)^2&-S_{n-2}(x)S_n(x)=\\ &=\left(S_{n-2}(x)+\frac{x^{n-1}}{(n-1)!}\right)^2-S_{n-2}(x)\left(S_{n-2}(x)+\frac{x^{n-1}}{(n-1)!}+\frac{x^{n}}{n!}\right)\\ &=S_{n-2}(x)\left(\frac{x^{n-1}}{(n-1)!}-\frac{x^{n}}{n!}\right)+\frac{x^{2n-2}}{(n-1)!^2}\\ &=\frac{x^{n-1}}{(n-1)!}S_{n-2}(x)\left(1-\frac{x}{n}\right)+\frac{x^{2n-2}}{(n-1)!^2}\\ &=\frac{x^{n-1}}{(n-1)!}\left(\sum_{m=0}^{n-2}\frac{x^m}{m!}\right)\left(1-\frac{x}{n}\right)+\frac{x^{2n-2}}{(n-1)!^2}\\ &=\frac{x^{n-1}}{(n-1)!}\left(\sum_{m=0}^{n-2}\frac{x^m}{m!}-\frac{1}{n}\sum_{m=0}^{n-2}\frac{x^{m+1}}{m!}\right) +\frac{x^{2n-2}}{(n-1)!^2}\\ &=\frac{x^{n-1}}{(n-1)!}\left(\sum_{m=0}^{n-2}\frac{x^m}{m!}-\frac{1}{n}\sum_{m=1}^{n-1}\frac{x^{m}}{(m-1)!}\right) +\frac{x^{2n-2}}{(n-1)!^2}\\ &=\frac{x^{n-1}}{(n-1)!}\left(1+\sum_{m=1}^{n-2}\left(\frac{1}{m}-\frac{1}{n}\right)\frac{x^m}{(m-1)!}-\frac{x^{n-1}}{n(n-2)!}\right) +\frac{x^{2n-2}}{(n-1)!^2}\\ &=\frac{x^{n-1}}{(n-1)!}\left(1+\sum_{m=1}^{n-2}\left(\frac{1}{m}-\frac{1}{n}\right)\frac{x^m}{(m-1)!}\right)\\ &\qquad\qquad+\frac{x^{2n-2}}{(n-1)!(n-2)!}\left(\frac{1}{n-1}-\frac{1}{n}\right)\tag{4}\\ &>0\\ \end{align*}

From the line (4) it's obvious that the inequality is valid and we may therefore conclude that $S_n$ is discrete log-concave.

The challenge is of course to sharpen the inequality (4) in order to obtain (3).

Note: Since the function $\ln$ is monotonically increasing we obtain due to the log-concavity of $S_n$

\begin{align*} \ln\left(S_{n-1}^2(x)\right)&\geq\ln\left(S_n(x)S_{n-2}(x)\right)\\ 2\ln\left(S_{n-1}(x)\right)&\geq\ln\left(S_n(x)\right)+\ln\left(S_{n-2}(x)\right)\\ \end{align*}

which may of some use for further calculations.

I can reduce the problem to a purely polynomial identity, without any logs in it. As explained by both @MarkusScheuer and @Arashium, the inequality to be shown is equivalent to $S_{n-1}^2\geq S_{n-2}S_n\frac{\ln(S_n)}{\ln(S_{n-1})}$. To isolate one of the logs, let us put $\phi_n=\frac{S_{n-1}^2}{S_{n-2}S_n}\ln(S_{n-1})-\ln(S_n)$. The goal is then to show that $\phi_n$ is nonnegative. Since $\phi_n(0)=0$, it will suffice to show that $\phi'_n$ is nonnegative. If we set $F_n=\frac{S_{n-1}^2}{S_{n-2}S_n}$, then

$$\phi'_n=F'_n\ln(S_{n-1})+F_n\frac{S_{n-2}}{S_{n-1}}-\frac{S_{n-1}}{S_{n}}= F'_n\ln(S_{n-1}) \tag{1} $$

Since $\ln(S_{n-1}) \geq 0$, it suffices to show that $F'_n$ is nonnegative. A little computation shows that

$$ F'_n=\frac{S_{n-1}}{(S_{n-2}S_n)^2}G_n, \ \text{with} \ G_n=2S_nS_{n-2}^2-S_{n-1}(S_{n-3}S_n+S_{n-2}S_{n-1}) $$

It will suffice to show that the rescaled polynomial $H_n=\frac{n!(n-1)!(n-2)!}{x^{n-2}}G_n$ satisfies

$$ H_n=\sum_{0 \leq i \leq j \leq n-2} (n-j)!(n-j-1)!(j-i)! \binom{n-2}{j}\binom{j}{i}\binom{2n+1-i-j}{j-i}x^{i+j} \tag{2} $$

I have checked that (2) is true for any $n\leq 40$ with the help of a computer, but failed to find a proof so far. I created a separate question for the proof of (2).

I'M SORRY, THIS ANSWER WAS BROKEN. In fact, $(\dagger)$ does not imply $(\ast)$. I'll leave it up for now, until I've figured out what best to do with it.

The point of this answer is to provide a reduction to a different polynomial inequality. Although I can't prove this inequality either, I am more optimistic about it, because it is quadratic in the $S_n$, instead of cubic like Ewan's, and because I can show it is logically equivalent to Ewan's inequality.

Let's recall that Ewan's inequality is $$-S_{n-1}^2 S_{n-2} + 2 S_n S_{n-2}^2 - S_{n} S_{n-1} S_{n-3} \geq 0 \quad (\ast).$$ The point of this answer is to show that $(\ast)$ is logically equivalent to $$\frac{n-1}{n} S_{n-1}^2 \leq S_n S_{n-2}. (\dagger)$$ For $2 \leq n \leq 30$, I've checked the stronger statement that the coefficients of $S_n S_{n-2} - \frac{n-1}{n} S_{n-1}^2$ are positive.

Proof that $(\ast)$ implies $(\dagger)$: Dividing by $S_n S_{n-1} S_{n-2}$, $(\ast)$ is $$0 \leq - \frac{S_{n-1}}{S_n} + 2 \frac{S_{n-2}}{S_{n-1}} - \frac{S_{n-3}}{S_{n-2}} = \frac{d}{dx} \left( \log \frac{S_{n-1}^2}{S_{n} S_{n-2}} \right)$$ So $(\ast)$ is equivalent to the claim that $\frac{S_{n-1}^2}{S_{n} S_{n-2}}$ is increasing. We note that $\lim_{x \to \infty} \frac{S_{n-1}^2}{S_{n} S_{n-2}} = \frac{n}{n-1}$ so, if $(\ast)$ holds, then $\frac{S_{n-1}^2}{S_{n} S_{n-2}} \leq \frac{n}{n-1}$ for all $x$, which rearranges to $(\dagger)$. $\square$

Incidentally, we can also use this to give another proof that $(\ast)$ is equivalent to the original inequality: $\frac{S_{n-1}^2}{S_{n} S_{n-2}} = \frac{S_{n-1}/S_{n}}{S_{n-2}/S_{n-1}}$. If $f(x)/g(x)$ is increasing, then so is $\int_0^y f(x) dx/\int_0^y g(x) dx$ (see, for example, [here][1]). So $\frac{\int S_{n-1}/S_{n}}{\int S_{n-2}/S_{n-1}} = \frac{\log S_{n}}{\log S_{n-1}}$ is increasing.

Proof that $(\dagger)$ implies $(\ast)$. Starting with $(\dagger)$, multiply by $S_n S_{n-2} x^{2n-2}/(n! (n-1)!)$ to get $$S_n S_{n-1}^2 S_{n-2} \frac{x^n x^{n-2}}{n! (n-2)!} \geq \frac{(x^{n-1})^2}{(n-1)!^2} S_n^2 S_{n-2}^2.$$ I miscopied the sign at this step; the inequality in the middle should point the other way. I've deleted the steps after this, but the idea was to get to do some algebra, get to an AM-GM, and then do some more algebra. Since the AM-GM is the only non-reversible step, this means we can't prove $(\ast)$ by this route.

In other words, $$(1/2) (x^n/n! S_{n-1} S_{n-2} + x^{n-2}/(n-2)! S_n S_{n-1}) \geq S_n (x^{n-1}/(n-1)!) S_{n-2} \quad (\ast \ast)$$ is presumably true, since it is logically equivalent to $(\ast)$. But $$\sqrt{x^n/n! S_{n-1} S_{n-2} \cdot x^{n-2}/(n-2)! S_n S_{n-1}} \geq S_n (x^{n-1}/(n-1)!) S_{n-2}$$ is false already for $n=2$. So any attempt to prove $(\ast \ast)$ must do something more gentle than AM-GM.