Does a transitive kernel action on fibers imply the domain is a group?

Question

Let $f:A\to B$ be a monoid homomorphism. Suppose the kernel acts transitively on every fiber. Does this imply $A$ is a group?

Certainly it implies every kernel element has a right inverse, which implies the kernel is a group, but I don't see how to get to the other fibers...

Answer 1

For a counterexample let $A = {\mathbb Z} \oplus {\mathbb Z}_{\ge 0}$, $B= {\mathbb Z}_{\ge 0}$, both under addition, and define $f$ by $f((a,b))=b$.

Then the kernel is $\{(a,0) : a \in {\mathbb Z}\}$, which is the group ${\mathbb Z}$, and it does indeed act transitively on the fibers $f^{-1}(i) = \{(a,i) : a \in {\mathbb Z}\}$.