Is there a "simple" form to \begin{align} \sum_{x=0}^{x_1}\frac{(A+2x)!}{(2x)!}\, \tag{1} \end{align}
Mathematica spits out the not very illuminating \begin{align} \sum_{x=0}^{x_1}\frac{(A+2x)!}{(2x)!}=2^{-A-2} A!-\frac{\left(A+2 x_1+2\right)! \, _3F_2\left(1,\frac{A}{2}+x_1+\frac{3}{2},\frac{A}{2}+x_1+2;x_1+\frac{3}{2},x_1+2;1\right)}{\left(2 \left(x_1+1\right)\right)!}\, . \end{align} Here, $x$, $x_1$ and $A$ are non-negative integers.
I suspect the cases where $A$ is even or odd probably have to be treated differently. Still according to Mathematica, the simpler expression $$ \sum_{x=0}^{x_0}\frac{(A+x)!}{x!}=\frac{(1+A+x_1)!}{(1+A)x_1!}$$ so one would think that the more complicated one should also eventually give something "simpler".
(Any reference or insight to this type of summation is welcomed.)
I don't think that there is a general answer.
Consider $$ I = \sum_{k=0}^{2x_1} \begin{pmatrix}A+k \\ k\end{pmatrix} $$ and $$ J = \sum_{k=0}^{2x_1} (-1)^k \begin{pmatrix}A+k \\ k\end{pmatrix} $$ What you look for is obviously $I+J$ (multiplied by $A!/2$) as odd terms cancel out and only even terms remain.
The first one is $I = \begin{pmatrix}A+2x_1 + 1\\ 2x_1\end{pmatrix}$ like you mentioned.
The later one seems trickier. You can maybe refer to negative binomial coefficients where binomial coefficients can be extended to $\begin{pmatrix}p\\ k\end{pmatrix}$ where $p\in\mathbb Z$. The definition is ( $n\in\mathbb N$) $$ \begin{pmatrix} -n \\ k \end{pmatrix} := (-1)^k \begin{pmatrix} n+k-1 \\ k \end{pmatrix} $$ With this convention, $$ J = \sum_{k=0}^{2x_1} (-1)^k \begin{pmatrix}A+k \\ k\end{pmatrix} = \sum_{k=0}^{2x_1} \begin{pmatrix}1-A \\ k\end{pmatrix} $$ and the latter has no closed forms when $1-A$ is positive, but it doubtfully has one when it is also negative which is your case.
There might be special cases like $A=1$ or $A=0$.