Codex @codex  0

Rearrange the identity intoTaking determinants gives $\det A\det (B-A-I)=1$. Hence $\det A\ne 0$, so $A$ is an invertible matrix, with $A^{-1}=B-A-I$. This uses that the matrices are finite and square: a right inverse then implies a two-sided inverse. Applying the identity to a vector in the kernel without first establishing commutation would not justify cancelling an $A$ on the wrong side. The argument is a case of invertibility forced by a polynomial product identity.