Finite Monoid with Right Cancellable Operation is Group

Theorem
Let $\left({S, \circ}\right)$ be a finite monoid.

Let $\circ$ be a right cancellable operation.

Then $\left({S, \circ}\right)$ is a group.

Proof
Group axioms $G0$, $G1$ and $G2$ are satisfied by dint of $\left({S, \circ}\right)$ being a monoid.

Recall the definition of right cancellable operation:
 * $\forall a, b, c \in S: a \circ c = b \circ c \implies a = b$

Let $\rho_c: S \to S$ be the right regular representation of $\left({S, \circ}\right)$ with respect to $c$.

By Right Cancellable iff Right Regular Representation Injective, $\rho_c$ is an injection.

By Regular Representation on Finite Structure is Bijection, $\rho_c$ is a bijection.

Thus $a \circ b = e$ has a unique solution for all $a \in S$.

That is, group axiom $G3$ holds on $S$.