This is proven in Inversion of Matrices over a Commutative Semiring by Reutenauer and Straubing.
The proofs aren't especially short though (as you request in the comments).
The first two paragraphs give good context for their paper though:
It is a well-known consequence of the elementary theory of vector spaces
that if $A$ and $B$ are $n$-by-$n$ matrices over a field (or even a skew field) such
that $AB = 1$, then $BA = 1$. This result remains true for matrices over a
commutative ring, however, it is not, in general, true for matrices over noncommutative
In this paper we show that if $A$ and $B$ are $n$-by-$n$ matrices over a
commutative semiring, then the equation $AB = 1$ implies $BA = 1$. We give
two proofs: one algebraic in nature, the other more combinatorial. Both
proofs use a generalization of the familiar product law for determinants over
a commutative semiring.
It's worth mentioning that "semi-ring" for them requires having a multiplicative identity, so they do prove the result you desire.