Inverse Completion of Commutative Semigroup is Inverse Completion of Itself

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

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 $x \circ' y^{-1}$ be cancellable for $\circ'$, where $x \in S$ and $y \in C$.

We have that $y$ is invertible for $\circ'$.

So by Invertible Element of Monoid is 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 Semigroup form Subsemigroup, $x$ is cancellable for $\circ$.

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

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.