I just had a thought, using the commutator matrix $C=AB-BA=1-BA$. Post-multiply by $B$ we get

$$ CB=B-BAB=B-B=0 $$

using $AB=1$. Now $B$ has full row rank and therefore $C=0$ implying that $A$ and $B$ commute. Is that a way to go or did I miss something?

I just had a thought, using the commutator matrix $C=AB-BA=1-BA$. Post-multiply by $B$ we get

$$ CB=B-BAB=B-B=0 $$

using $AB=1$. Now $B$ has full row rank and therefore $C=0$ implying that $A$ and $B$ commute. Is that a way to go or did I miss something?

EDIT:

Better yet, just consider the matrix $S=ABA$. $$ S=(AB)A=A $$ on the other hand $$ S=A(BA) $$ Therefore $$ A=A(BA) \Rightarrow A(BA-1)=0 \Rightarrow BA=1 $$