I have obtained a formula for the generating function of your sequence.

Let $S_n$ be defined as in the quesion. We extend the definition to $n = 0$ by demanding $0^0 = 0$, hence $S_0 = 0$.

Consider $S(t) = \sum_{n\geq 0} S_n t^n$. I will work out a formula for $S(t)$.

\begin{eqnarray*}
S(t) &=& \sum_{n=0}^\infty \sum_{j = 0}^n\frac{(-e)^{-j}}{j!}(n - j)^j t^n\\
&=& \sum_{j = 0}^\infty \frac{(-e)^{-j}}{j!}t^j\sum_{n = 0}^\infty n^jt^n\\
&=& \sum_{j = 0}^\infty \frac{(-e^{-1}t)^j}{j!}\frac{tA_j(t)}{(1 - t)^{j + 1}}\\
&=& \frac{t}{1 - t}\sum_{j = 0}^\infty A_j(t)\frac{(\frac{e^{-1}t}{t - 1})^j}{j!}\\
&=& \frac{t}{1 - t}\frac{t - 1}{t - e^{e^{-1}t}}\\
&=& \frac{t}{e^{e^{-1}t} - t}.
\end{eqnarray*}

Explanation for the calculation:

The key step is $\sum_{n\geq 0}  n^jt^n = \frac{tA_j(t)}{(1 - t)^{j + 1}}$, where $A_j(t)$ is the [Eulerian polynomial][1]. We then use the generating function $\sum_{j \geq 0}A_j(t)\frac{x^j}{j!} = \frac{t - 1}{t - e^{(t - 1)x}}$.

  [1]: https://en.wikipedia.org/wiki/Eulerian_number