$$k_a(x)=\sum^\infty_{n=0}{\frac{n^2}{(n+x)!}}$$ $$k_b(x)=\sum^\infty_{n=0}{\frac{(n+x)^2}{n!}}$$ I noticed these functions closely relate to $e$. By looking at them I was able to determain a closed from for $k_b$. $$k_b(x)=e(x^2+2x+2)$$ Though I still have not found a short form for $k_a$. I do think it exists. Some integer inputs are as follows.
$k_a(0)=2e$
$k_a(1)=e-1$
$k_a(2)=2e-5$
$k_a(3)=5e-\frac{27}{2}$
Can any of you take a stab at it?
On
This is too long for a comment.
Considering $$k_a(x)=\sum^\infty_{n=0}{\frac{n^2}{(n+x)!}}$$ and using the result from Wolfram Alpha, it seems that the expression could simplify to $$k_a(x)=\frac{(2+x-x^2)+e\,(2-2x+x^2)\left(\Gamma (x+2)-(x+1)\, \Gamma (x+1,1) \right) }{\Gamma (x+2)}$$ which, as user404188 noticed, for integer values of $x$, reduces to $$k_a(x)=(x^2-2x+2)\,e + \mathrm{rational}$$ as shown in the table. $$\left( \begin{array}{cc} x & k_a(x) \\ 0 & 2 e \\ 1 & -1+e \\ 2 & -5+2 e \\ 3 & -\frac{27}{2}+5 e \\ 4 & -\frac{163}{6}+10 e \\ 5 & -\frac{1109}{24}+17 e \\ 6 & -\frac{2827}{40}+26 e \\ 7 & -\frac{14483}{144}+37 e \\ 8 & -\frac{685007}{5040}+50 e \\ 9 & -\frac{2374691}{13440}+65 e \\ 10 & -\frac{80885629}{362880}+82 e \end{array} \right)$$
The expression can rewrite $$k_a(x)=e \left(x^2-2 x+2\right)-\frac{x^2-x-2}{\Gamma (x+2)}-e \left(x^3-x^2+2\right)\frac{ \Gamma (x+1,1)}{\Gamma (x+2)}$$
On
We have the Mittag-Leffler function defined by
$$E_{\alpha,\beta}(z)\equiv\sum_{k=0}^\infty\frac{z^k}{\Gamma(\alpha k+\beta)}$$
In particular,
$$E_{1,x+1}(z)=\sum_{k=0}^\infty\frac{z^k}{(k+x)!}$$
$$z\frac\partial{\partial z}E_{1,x+1}(z)=\sum_{k=0}^\infty\frac{kz^k}{(k+x)!}$$
$$z\frac\partial{\partial z}z\frac\partial{\partial z}E_{1,x+1}(z)=\sum_{k=0}^\infty\frac{k^2z^k}{(k+x)!}$$
And thus your sum can compactly be written as
$$S_a(x)=\lim_{z\to1}\frac\partial{\partial z}z\frac\partial{\partial z}E_{1,x+1}(z)$$
$$k_a(x)=\sum^\infty_{n=0}{\frac{n^2}{(n+x)!}}$$ In order to extend from $x$ integer to real : $(n+x)!=\Gamma(n+x+1)$ $$k_a(x)=\sum^\infty_{n=0}{\frac{n^2}{\Gamma(n+x+1)}}$$ Let : $\quad y(t)=\sum^\infty_{n=0}{\frac{t^n}{\Gamma(n+x+1)}}=e^tt^{-x}\left(1-\frac{\Gamma(x,t)}{\Gamma(x)} \right)$ $$\frac{dy}{dt}=\sum^\infty_{n=0}{\frac{nt^{n-1}}{\Gamma(n+x+1)}}=\left(e^tt^{-x}-xe^tt^{-x-1} \right)\left(1-\frac{\Gamma(x,t)}{\Gamma(x)} \right)- e^tt^{-x}\frac{1}{\Gamma(x)}\left(-e^{-t} t^{x-1} \right)$$ $$\frac{dy}{dt}=\sum^\infty_{n=0}{\frac{nt^{n-1}}{\Gamma(n+x+1)}}=e^t t^{-x-1}(t-x)\left(1-\frac{\Gamma(x,t)}{\Gamma(x)} \right)+\frac{t^{-1}}{\Gamma(x)}$$ $$g(t)=\sum^\infty_{n=0}{\frac{nt^{n}}{\Gamma(n+x+1)}}=e^tt^{-x}(t-x)\left(1-\frac{\Gamma(x,t)}{\Gamma(x)} \right)+\frac{1}{\Gamma(x)}$$
$$\frac{dg}{dt}=\sum^\infty_{n=0}{\frac{n^2t^{n-1}}{\Gamma(n+x+1)}}=e^t t^{-x-1}\left(x^2-2xt+t+t^2 \right)\left(1-\frac{\Gamma(x,t)}{\Gamma(x)} \right)-e^tt^{-x}(t-x)\frac{1}{\Gamma(x)}\left(-e^{-t} t^{x-1} \right)$$
For $t=1$ $$k_a(x)=\sum^\infty_{n=0}{\frac{n^2}{\Gamma(n+x+1)}}=e\,\left( x^2-2x+2\right)\left(1-\frac{\Gamma(x,1)}{\Gamma(x)} \right)+\frac{1-x}{\Gamma(x)}$$
Numerical tests are in agreement with this formula.
In addition :
CASE OF $x$ INTEGER $\qquad x=N$
$$\Gamma(N,1)=e^{-1}(N-1)!\:e_{_{N-1}}(1)=e^{-1}(N-1)!\sum_{k=0}^{N-1}\frac{1}{k!}$$ $(N-1)!\sum_{k=0}^{N-1}\frac{1}{k!}\quad$ is an integer. As a consequence $\quad e\,\frac{\Gamma(N,1)}{\Gamma(N)}$ is rational.
$$k_a(N)= \quad \begin{cases} \text{Irrational part }=(N^2-2N+2)\:e \\ \text{Rational part }= -\frac{N-1}{(N-1)!}-(N^2-2N+2)\displaystyle\sum_{k=0}^{N-1}\frac{1}{k!} \qquad\qquad N>0\end{cases}$$ This is in full agreement with the numerical table given by Claude Leibovici.