I think this question could be answered by only using mathematics (it relates to physics). Where, $\partial_{x}f$ is denoted as partial derivative of $f$ w.r.t $x$, and first exponential term behaves like operator at some sense.
My idea is expanding both sides into power series form (Taylor series).
LHS is
$$\sum_{n=0}^{\infty}\left(\frac{-\hbar}{2}\right)^{n}\frac{1}{n!}\partial_{a}^{n}\partial_{b}^{n}\sum_{l=0}^{\infty}\left(\frac{-ab}{\hbar}\right)^{l}\frac{1}{l!}$$
Expand the first sum to calculate derivative, it will be:
$$\sum_{k=0}^{\infty}\frac{1}{2^{k}k!}\sum_{l=0}^{\infty}\left(\frac{-ab}{\hbar}\right)^{l}\frac{(l+k)!}{(l!)^{2}}$$
While, RHS is
$$2\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{-2ab}{\hbar}\right)^n$$
And I don't see if it is right. Or, the considered equation is not even right. If the equation is correct (by mean "equation"), is there any way to prove it?
Originally, this equation is from a physics paper by A. C. Hirshfeld
It is equation 5.20. In the paper, $a=a$ and $\bar{a}=b$.
EDIT Below is what I have done: $$\sum_{n}^{\infty}\left(\frac{-\hbar}{2}\right)^{n}\frac{1}{n!}\partial_{a}^{n}\partial_{b}^{n}\sum_{l=0}^{\infty}\left(\frac{-ab}{\hbar}\right)^{l}\frac{1}{l!}=\sum_{l=0}^{\infty}\left(\frac{-ab}{\hbar}\right)^{l}\frac{1}{l!}+\frac{-h}{2}\partial_{a}\partial_{b}\sum_{l=0}^{\infty}\left(\frac{-ab}{\hbar}\right)^{l}\frac{1}{l!}+\\ +\left(\frac{-\hbar}{2}\right)^{2}\partial_{a}^{2}\partial_{b}^{2}\frac{1}{2!}\sum_{l=0}^{\infty}\left(\frac{-ab}{\hbar}\right)^{l}\frac{1}{l!}+...$$ $$=\sum_{l=0}^{\infty}\left(\frac{-ab}{\hbar}\right)^{l}\frac{1}{l!}+\frac{-\hbar}{2}\sum_{l=1}^{\infty}\left(\frac{-1}{\hbar}\right)^{l}a^{l-1}b^{l-1}\frac{l^{2}}{l!}+\\ +\left(\frac{-\hbar}{2}\right)^{2}\frac{1}{2!}\sum_{l=2}^{\infty}\left(\frac{-1}{\hbar}\right)^{l}a^{l-2}b^{l-2}\frac{l^{2}(l-1)^2}{l!}+...$$ $$=\sum_{l=0}^{\infty}\left(\frac{-ab}{\hbar}\right)^{l}\frac{1}{l!}+\frac{-\hbar}{2}\sum_{l=1}^{\infty}\left(\frac{-1}{\hbar}\right)\left(\frac{-1}{\hbar}\right)^{l-1}a^{l-1}b^{l-1}\frac{l}{(l-1)!}+\\ +\left(\frac{-\hbar}{2}\right)^{2}\frac{1}{2!}\sum_{l=2}^{\infty}\left(\frac{-1}{\hbar}\right)^{2}\left(\frac{-1}{\hbar}\right)^{l-2}a^{l-2}b^{l-2}\frac{l(l-1)}{(l-2)!}+...$$ $$=\sum_{k=0}^{\infty}\frac{1}{2^{k}}\frac{1}{k!}\sum_{l=0}^{\infty}\left(\frac{-1}{\hbar}\right)^{l}a^{l}b^{l}\frac{(l+k)!}{l!^{2}}$$