Inverse Completion of Commutative Semigroup is Inverse Completion of Itself

Theorem
Let $$\left({S, \circ}\right)$$ be a commutative monoid whose identity is $$e$$.

Let $$\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$$ be the subsemigroup of cancellable elements of $$\left({S, \circ}\right)$$.

Let $$\left({T, \circ'}\right)$$ be an inverse completion of $$\left({S, \circ}\right)$$.

Then $$\left({T, \circ'}\right)$$ is its own inverse completion.

Proof
Let $$x \circ' y$$ be cancellable for $$\circ'$$, where $$x \in S$$ and $$y \in C$$.

By definition, $$y$$ is invertible for $$\circ'$$.

By Invertible also Cancellable, $$y$$ is also cancellable for $$\circ'$$.

Now $$x = \left({x \circ' y^{-1}}\right) \circ y$$ by definition of inverse element.

Thus $$x$$ is also cancellable for $$\circ'$$, and by Cancellable Elements of a Semigroup, $$x$$ is cancellable for $$\circ$$.

So $$x \in S \Longrightarrow x \in C$$.

Thus $$x$$ is invertible for $$\circ'$$.

Hence $$x \circ' y$$ is invertible for $$\circ'$$ by Inverse of Product.

So every cancellable element of $$\left({T, \circ'}\right)$$ is invertible, and so $$\left({T, \circ'}\right)$$ is its own inverse completion.