Closed-form expression of sum of series $sumlimits_{n=0}^inftysumlimits_{k=1}^inftyfrac{x^k}{(k+2n)!}$

I would like a closed-form expression of the following series:

$\left(\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots\right)+\left(\frac{x}{3!}+\frac{x^2}{4!}+\frac{x^3}{5!}+\frac{x^4}{6!}+\cdots\right)+\left(\frac{x}{5!}+\frac{x^2}{6!}+\frac{x^3}{7!}+\frac{x^4}{8!}+\cdots\right)+\cdots$

Clearly the first bracket is $e^x-1$, the second bracket is $x^{-2}(e^x-1-x-\frac{x^2}{2!})$, the third bracket is $x^{-4}(e^x-1-x-\frac{x^2}{2!}-\frac{x^3}{3!}-\frac{x^4}{4!})$, and so on. Is there a way to combine these infinitely many terms into a closed form expression?

Answer

For any formal Laurent series in $t$, $f(t) = \sum\limits_{\ell=-\infty}^\infty \alpha_\ell t^\ell$, we will use the notation $[t^\ell] f(t)$ to denote the coefficient $\alpha_\ell$ in front of the monomial $t^\ell$. When $f(t)$ defines a
function holomorphic over an annular region $\mathcal{A} \subset \mathbb{C}$, $[t^\ell] f(t)$ can be reinterpreted as a contour integral over some suitably chosen circle $C_R = \{ t : |t| = R \}$ lying within $\mathcal{A}$.

$$[t^\ell]f(t) \quad\longleftrightarrow\quad\frac{1}{2\pi i}\oint_{C_R} \frac{f(t)}{t^{\ell+1}} dt$$

The series at hand can be rewritten as

$$\begin{align} \mathcal{S} \stackrel{def}{=}\sum_{\ell=0}^\infty \sum_{k=1}^\infty \frac{x^k}{(k+2\ell)!} &= \sum_{\ell=0}^\infty \sum_{k=1}^\infty \sum_{m=0}^\infty \delta_{m,k+2\ell} \frac{x^k}{m!} = \sum_{\ell=0}^\infty \sum_{k=1}^\infty \sum_{m=0}^\infty \delta_{m-k,2\ell} \frac{x^k}{m!}\\ &= \sum_{\ell=0}^\infty \left\{ \sum_{k=1}^\infty \sum_{m=0}^\infty [t^{2\ell}]\left(\frac{x}{t}\right)^k \frac{t^m}{m!}\right\}\tag{*1} \end{align}$$
Over the complex plane, the sum
$\displaystyle\;\sum\limits_{k=1}^\infty \sum\limits_{m=0}^\infty \left(\frac{x}{t}\right)^k \frac{t^m}{m!}\;$
converges absolutely for $|t| > |x|$.
If we interpret the expression inside curly braces of $(*1)$ as contour integrals over circle $C_R$ with $R > |x|$, we can switch the order of summation and $[\ldots]$.

We find

$$\mathcal{S} = \sum_{\ell=0}^\infty [t^{2\ell}] \sum\limits_{k=1}^\infty \sum\limits_{m=0}^\infty \left(\frac{x}{t}\right)^k \frac{t^m}{m!} = \sum_{\ell=0}^\infty [t^{2\ell}] \frac{xe^t}{t-x} = \sum_{\ell=0}^\infty [t^0] t^{-2\ell} \frac{xe^t}{t-x}$$

Over the complex plane, the term $\sum\limits_{\ell=0}^\infty t^{-2\ell}$ converges absolutely when $|t| > 1$. If we choose $R > \max\{ |x|, 1 \}$,
we can change the order of summation and $[\cdots]$ again and get

$$\mathcal{S} = [t^0]\left[\left(\sum_{\ell=0}^\infty t^{-2\ell}\right)\frac{xe^t}{t-x}\right] = [t^0] \left(\frac{t^2}{t^2-1}\frac{xe^t}{t-x}\right) = \frac{1}{2\pi i}\oint_{C_R} \frac{tx e^t}{(t^2-1)(t-x)} dt$$
Since $R > \max\{ |x|, 1 \}$, there are 3 poles $x, \pm 1$ inside $C_R$. If one
sums over the contributions from these poles, one obtains

$$\begin{align} \mathcal{S} &= \frac{x^2 e^x}{(x^2-1)} + \frac{xe}{2(1-x)} + \frac{-xe^{-1}}{(-2)(-1-x)}\\ &= \frac{x}{2(x^2-1)}\left[ 2xe^x - e (x+1) - e^{-1}(x-1)\right]\\ &= \frac{x}{2(x^2-1)}\left[ x(2e^x - (e+e^{-1})) - (e-e^{-1})\right] \end{align}$$
Reproducing what Did mentioned in hir comment.