Finite Monoid with Left Cancellable Operation is Group

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

Let $\circ$ be a left 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 left cancellable operation:
 * $\forall a, b, c \in S: c \circ a = c \circ b \implies a = b$

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

By Left Cancellable iff Left Regular Representation Injective, $\lambda_c$ is an injection.

By Left Regular Representation wrt Left Cancellable Element on Finite Semigroup is Bijection, $\lambda_c$ is a bijection.

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

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