Inverse Completion of Commutative Semigroup is Abelian Group

Let $$\left({S, \circ}\right)$$ be a commutative monoid, all of whose elements are cancellable.

Then an inverse completion of $$\left({S, \circ}\right)$$ is an abelian group.

Proof
Let $$\left({S, \circ}\right)$$ be a commutative monoid, all of whose elements are cancellable

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

From Inverse Completion Commutative Semigroup, $$T$$ is commutative.

The product of two invertible elements is invertible.

Every element of $$S \circ' S^{-1}$$ is invertible in $$T$$, from the definition of inverse completion and the fact that every element of $$S$$ is cancellable.

Thus every element of $$T$$ is invertible.

As Invertible Cancellable Elements of Monoid form Group, $$T$$ is therefore a group.

So $$\left({T, \circ'}\right)$$ is commutative, and a group, and therefore an abelian group.