(asked previously in MSE here)

In the course of a calculation, I have met the following complicated identity. Let $A$ and $a$ be positive integers. Then I believe that $$ \sum_{B\ge A,b\ge a} (-1)^{a+b}{b\choose a} {B-1\choose A-1}\frac{(B-1)!b!}{(B+b)(x-b)^{(B+b)}}=\frac{(A-1)!a!}{(A+a)(x-A+1)^{(A+a)}},$$ where $x$ is some variable and $(x)^{(a)}=x(x+1)\cdots(x+a-1)$ is the rising factorial. This is what I would like to prove.

Notice how the left hand side in principle has poles at all integer values of $x$, while the right hand side only has poles at integers smaller than $A-1$.