I want to find the closed form of $$ G_n(k) = \sum_{k=0}^n k! \bigg\lbrace {n \atop k}\bigg\rbrace x^k $$
Suppose the closed form of $$ E_n(k) = \sum_{k=0}^n k!\, x^k $$ is known; call it $ A_n(k)$. And suppose also that the closed form of $$ F_n(k) = \sum_{k=0}^n \bigg\lbrace {n \atop k}\bigg\rbrace x^k $$ is known; call it $B_n(k) $.
Using these generating functions and their closed forms, can I acquire a closed form for $G_n(k)$ ? I know I can't just multiply the two functions because of the powers of $x$, but I don't think a typical convolution will give a congenial result.
Your series above can be re-written as (infinite) power series
\begin{align*} e_{n}(x) & = \sum_{k=0}^\infty k! \mathbf{1}_{k \leq n} x^k \\ f_n(x) & = \sum_{k=0}^\infty \bigg\lbrace {n \atop k}\bigg\rbrace \mathbf{1}_{k \leq n} x^k \\ g_n(x) & = \sum_{k=0}^\infty k! \bigg\lbrace {n \atop k}\bigg\rbrace \mathbf{1}_{k \leq n} x^k \\ & = \sum_{k=0}^\infty\bigg(k!\, \mathbf{1}_{k \leq n}\bigg)\bigg( \bigg\lbrace {n \atop k}\bigg\rbrace \mathbf{1}_{k \leq n} \bigg)x^k \end{align*}
where we recognise the last line as the Hadamard product of the power series $e_n, \, f_n$ (i.e. the term-wise product).
According to the Wikipedia page above, if closed forms are known for $e_n, f_n$ (equivalently $E_n, \, F_n$) then
$$ G_n(x) = g_n(x) =\frac1{2\pi} \int_{0}^{2\pi} e_n\big(\sqrt{z} e^{zt}\big) \, f_n\big(\sqrt{z} e^{-zt}\big) dt$$