Left-invertible elements of monoid form submonoid

Statement
Let $$M$$ be a monoid with neutral element $$e$$. Then, the set of left-invertible elements of $$M$$ form a submonoid of $$M$$.

Related facts
Inverse map is involutive: Specifically the reversal law for inverses.

Proof
Given: A monoid $$M$$ with neutral element $$e$$

To prove: $$e$$ is left-invertible, and further, if $$a_1, a_2$$ are left-invertible elements of $$M$$, so is $$a_1 * a_2$$

Proof: Clearly, $$e$$ is left-invertible, since $$e * e = e$$.

Suppose $$a_1,a_2$$ are left-invertible and $$b_1,b_2$$ are left inverses for $$a_1, a_2$$. Then consider:

$$(b_2 * b_1) * (a_1 * a_2) = ((b_2 * b_1) * a_1) * a_2 = (b_2 * (b_1 * a_1)) * a_2 = (b_2 * e) * a_2 = b_2 * a_2 = e$$

Thus, $$b_2 * b_1$$ is a left inverse for $$a_1 * a_2$$, so $$a_1 * a_2$$ is left-invertible, completing the proof.