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*}