If A, B and $\big(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\big)$ are non-singular matrices of the same order, then the inverse of $A\Big(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\Big)^{-1}B$ is equal to:

Use identity: $\operatorname{adj}(M^{-1})=\dfrac{\operatorname{adj}(M)}{|M|}$

So,
$\operatorname{adj}(A^{-1})=\dfrac{A}{|A|},\quad \operatorname{adj}(B^{-1})=\dfrac{B}{|B|}$

Hence,
$\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})=\dfrac{A}{|A|}+\dfrac{B}{|B|}$

Given expression:
$X=A\Big(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\Big)^{-1}B$

$X^{-1}=B^{-1}\Big(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\Big)A^{-1}$

Substitute:
$=B^{-1}\left(\dfrac{A}{|A|}+\dfrac{B}{|B|}\right)A^{-1}$

$=\dfrac{1}{|A|}B^{-1}AA^{-1}+\dfrac{1}{|B|}B^{-1}BA^{-1}$

$=\dfrac{1}{|A|}B^{-1}+\dfrac{1}{|B|}A^{-1}$

$=\dfrac{1}{|A|} \dfrac{adjB}{|B|}$+$\dfrac{1}{|B|} \dfrac{adjA}{|A|}$
$=\dfrac{adjB}{|AB|}$+$\dfrac{adjA}{|AB|}$