Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup

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 $$T = S \circ' C^{-1}$$, and is a commutative semigroup.

Proof
Let $$x, z \in S, y, w \in C$$. Then by Associativity and Commutativity Properties, $$x, y, z, w$$ all commute with each other under $$\circ$$.

Thus $$\left({x \circ z}\right)\circ' {\left({w \circ y}\right)^{-1} \in S \circ' C^{-1}$$, proving that $$S \circ' C^{-1}$$ is closed, therefore a subsemigroup of $$\left({T, \circ'}\right)$$.

It is also easily shown (WIP) that $$x \circ' y^{-1}$$ commutes with $$z \circ' w^{-1}$$, so $$S \circ' C^{-1}$$ is shown to be a commutative subsemigroup of $$\left({T, \circ'}\right)$$.