In a monoid, a left inverse and a right inverse of the same element are equal; hence an invertible element has exactly one inverse, and it is two-sided

Since $e$ is a right identity and $g * h' = e$, we may write $h = h * e = h * (g * h')$.

Since $e$ is a left identity and $h * g = e$, we may write $h' = e * h' = (h * g) * h'$.

Associativity identifies the two right-hand sides: $h * (g * h') = (h * g) * h'$.

Combining, $h = h * (g * h') = (h * g) * h' = h'$, which is the stated equality of a left inverse and a right inverse.

Claim 3: if $h$ is a left inverse and $h'$ a right inverse of $g$, then $h = h'$ by step 2.1, so this common element satisfies both $h * g = e$ and $g * h = e$ and is a two-sided inverse; hence $g$ is invertible.

Claim 1: let $u$ and $v$ be two-sided inverses of $g$. Then $u$ is a left inverse and $v$ is a right inverse of $g$, so $u = v$ by step 2.1; the two-sided inverse of an invertible element is therefore unique, and the notation $g^{-1}$ is legitimate.

Claim 2: let $g$ be invertible with two-sided inverse $g^{-1}$. If $h * g = e$ then $h$ is a left inverse and $g^{-1}$ is a right inverse of $g$, so $h = g^{-1}$ by step 2.1; if $g * h = e$ then $g^{-1}$ is a left inverse and $h$ a right inverse, so $g^{-1} = h$ by step 2.1.

A left inverse and a right inverse of the same element coincide, an invertible element has exactly one inverse, and either one-sided equation already identifies it.