Proof of $\sum\limits_{n=1}^{\infty} \frac{x^n \log(n!)}{n!} \sim x \log(x) e^x$ as $x \to \infty$
I will try to show the first asymptotics.
We begin with the following quantitative form of Stirling's formula:
Fact. For all $n \geq 0$, $$ \log (n!) = (n + \tfrac{1}{2})\log(n+1) - n + \mathcal{O}(1). \tag{1} $$
Now let $N_t$ be a Poisson random variable of rate $t$. Then
\begin{align*} \smash[b]{\sum_{n=0}^{\infty} \frac{t^n \log (n!)}{n!}e^{-t}} &= \Bbb{E}[\log (N_t !)] \\ &= \Bbb{E}[N_t \log (N_t) + \tfrac{1}{2}\log (N_t + 1) - N_t + \mathcal{O}(1)] \\ &= \Bbb{E}[N_t \log (N_t + 1)] + \tfrac{1}{2}\Bbb{E}[\log(N_t + 1)] - t + \mathcal{O}(1). \tag{2} \end{align*}
Now we claim the following:
Claim. For any $a \geq 0$ we have $$ t\log(t+a) \leq \Bbb{E}[N_t \log(N_t + a)] = t \Bbb{E}[\log(N_t + a + 1)] \leq t \log(t+ a + 1). \tag{3} $$ Here, we use the convention that $0 \log 0 = 0$.
Assuming this claim, we easily find that
$$ \Bbb{E}[N_t \log(N_t + 1)] = t \log t + \mathcal{O}(1) \quad \text{and} \quad \Bbb{E}[\log(N_t + 1)] = \log t + \mathcal{O}(t^{-1}). $$
Plugging this to $\text{(2)}$ gives
$$ \sum_{n=0}^{\infty} \frac{t^n \log (n!)}{n!}e^{-t} = (t + \tfrac{1}{2})\log t - t + \mathcal{O}(1) = \log (t!) + \mathcal{O}(1). $$
Dividing both sides by $t \log t$ yields the first asymptotics.
Proof of Claim. The last inequality of $\text{(3)}$ is easy to prove. Since the function $x \mapsto \log(x+a+1)$ is concave, by the Jensen's inequality we have
$$ \Bbb{E}[\log(N_t + a + 1)] \leq \log(\Bbb{E} N_t + a + 1) = \log(t+a+1). $$
In order to show the first inequality of $\text{(3)}$, notice that $x \mapsto x\log(x+a)$ is convex (with the 2nd derivative $(2a+x)/(a+x)^2 > 0$). Thus by the Jensen's inequality again
$$ \Bbb{E}[N_t \log (N_t + a)] \geq (\Bbb{E}N_t) \log (\Bbb{E}N_t + a) = t \log(t+a). $$
Finally, the middle equality of $\text{(3)}$ is given by
\begin{align*} \Bbb{E}[N_t \log (N_t + a)] &= \sum_{n=1}^{\infty} n \log(n+a) \cdot \frac{t^n}{n!}e^{-t} \\ &= \sum_{n=0}^{\infty} \log(n+a+1) \cdot \frac{t^{n+1}}{n!}e^{-t} = t \Bbb{E}[\log (N_t + a + 1)]. \end{align*}
Your idea is right that the entries near $n=x$ matter the most. In particular, the entries with $n=x+O(\sqrt{x})$ dominate this sum.
The following is not a complete answer, but I believe the holes can be filled in.
Suppose that $N=x + \alpha \sqrt{x},$ where $\alpha$ is fixed. Then, by Stirling's formula \begin{align*} \frac{\ln(N!)}{N!} x^N &= \left(1+O\left(\frac{1}{\ln x}\right)\right) \frac{(x+\alpha \sqrt{x}) \ln x}{\sqrt{2 \pi x} (x+\alpha \sqrt{x})^{x+\alpha \sqrt{x}}}e^{x+\alpha \sqrt{x}} x^{x+\alpha \sqrt{x}} \\ &=\left(1+O\left(\frac{1}{\ln x}\right)\right) x (\ln x) e^x \left[e^{\alpha \sqrt{x}} \left( \frac{x}{x+\alpha \sqrt{x}}\right)^{x+\alpha \sqrt{x}} \frac{1}{\sqrt{2\pi x}} \right]. \end{align*}
Now, note that \begin{align*} \left( \frac{x}{x+\alpha \sqrt{x}}\right)^{x+\alpha \sqrt{x}}&=\exp \left( -(x+\alpha \sqrt{x}) \cdot \ln\left(1+\frac{\alpha}{\sqrt{x}} \right)\right) \\&=\exp\left( -(x+\alpha \sqrt{x}) \left( \frac{\alpha}{\sqrt{x}} - \frac{\alpha^2}{2x} + O(n^{-3/2}) \right) \right) \\&= \exp \left( -\alpha \sqrt{x} - \alpha^2/2 + O(x^{-1/2}) \right). \end{align*}
Therefore $$ \frac{\ln N!}{N!}x^N = \left(1+O\left(\frac{1}{\ln x}\right)\right) [x \ln(x) e^x] \frac{e^{-\alpha^2/2}}{\sqrt{2\pi}} \frac{1}{\sqrt{x}} $$
Now the sum of these $e^{-\alpha^2/2}{\sqrt{2\pi}} \frac{1}{\sqrt{x}}$ over $N \in [x-C \sqrt{x}, x+C \sqrt{x}]$, $(C$ fixed w.r.t. $x$), is a Riemann sum for the integral $\int_{-C}^C \frac{e^{-t^2/2}}{\sqrt{2\pi}}dt$. This integral can be arbitrarily close to 1 for $C$ large enough.
It remains to be shown that for any $\epsilon >0$ (fixed w.r.t. $x$), $$ \sum_{n \notin [x-C \sqrt{x}, x+C \sqrt{x}]} \frac{\ln(N!)}{N!} x^N < \epsilon \, x \ (\ln x) \ e^x, $$ for $C$ large enough.