We write $G_n(t)=\sum_{k=0}^n S(n,k)k!t^k$. This agrees with your definition when $n\geq 1$. For $n=0$, we have $G_0(t)=1$. If $t$ is a positive integer, then $G_n(t)$ is the number of ways to partition $[n]=\{1,2,\ldots, n\}$ into $k\leq n$ parts and give each part a colored label, where there are $t$ available colors and $k$ labels. This $G_n(t)$ is called Fubini polynomial. We have an exponential generating function of Stirling numbers of the second kind:
$$
\sum_{n=k}^{\infty} S(n,k)k! \frac{x^n}{n!} = (e^x-1)^k.
$$
Applying this, we may find a bivariate generating function:
$$
\sum_{k=0}^{\infty}\sum_{n=k}^{\infty} S(n,k)k! t^k\frac{x^n}{n!}=\sum_{k=0}^{\infty} (e^x-1)^k t^k = \frac1{1-t(e^x-1)}.
$$
Thus, by switching the order of summation, we have
$$
\sum_{n=0}^{\infty} G_n(t)\frac{x^n}{n!} = \frac1{1-t(e^x-1)}
$$
with the absolute convergence is provided when $|t(e^x-1)|<1$.
The polynomial $G_n(t)$ enjoys a few nice properties.
We have $G_n(-1)=(-1)^n$. This can be proved through a polynomial identity for Stirling numbers (Theorem 0.1 in my note below). This also has a beautiful combinatorial proof by Grigory M here
By the recurrence relation $S(n+1,k)=kS(n,k)+S(n,k-1)$, we have
$$
S(n+1,k)k!=k\big(S(n,k)k!+S(n,k-1)(k-1)!\big).$$
This gives the following recurrence of $G_n(t)$:
$$
G_{n+1}(t)=tG_n'(t)+t\big(tG_n(t)\big)'
= t\big(G_n(t)+(t+1)G_n'(t)\big) $$
$$
=t\big((t+1)G_n(t)\big)'.
$$
Using $G_1(t)=t$ has one real root, we can prove by the above recurrence, Rolle's theorem, and induction on $n$ that $G_n(t)=0$ has $n$ distinct real root for $n\geq 1$. Of course one of the real root is zero, the other roots are negative.
Moreover, by Newton's inequality, the coefficient sequence of $G_n(t)$ is log-concave, and hence unimodal.
There is a symmetry for $F_n(t)=G_n(t)/t$ about $t=-1/2$. We have for $n\geq 1$,
$$F_n(t)=(-1)^{n-1}F_n(-t-1).$$
I wrote proof of this identity (R. Stanley "Enumerative Combinatorics" volume 1, Chapter 3-Exercise 141(d)) here: http://www.csun.edu/~sungjin/SJK_StirlingS2.pdf
There is a combinatorial proof relying solely on $G_n(-1)=(-1)^n$. This is done by first proving
$$
S(n,r+1)(r+1)!=\sum_{k=r+1}^n S(n,k)k!(-1)^{n-k}\binom{k-1}{r}
$$
considering the identity $G_m(-1)=(-1)^m$ for each partitioned blocks of $[n]$ whose size is $m$.
Therefore, the distinct $n$ real roots of $G_n(t)$ except for $0$ must be symmetric about $t=-1/2$. Since $F_n(t)$ is increasing for $t\geq 0$, there are no roots in $t\leq -1$. Hence, all real roots of $F_n(t)$ are distinct and in $(-1,0)$ interval.
