The invertible elements of a monoid form a group under the restricted operation

$e\in M^{\times }$, since $e\ast e=e$ exhibits $e$ as a two-sided inverse of itself.

Let $g,h\in M^{\times }$ with inverses $g^{-1},h^{-1}$. Then $(g\ast h)\ast (h^{-1}\ast g^{-1})=g\ast (h\ast h^{-1})\ast g^{-1}=g\ast e\ast g^{-1}=g\ast g^{-1}=e$ and $(h^{-1}\ast g^{-1})\ast (g\ast h)=h^{-1}\ast (g^{-1}\ast g)\ast h=h^{-1}\ast e\ast h=h^{-1}\ast h=e$, the regroupings being licensed by associativity. So $h^{-1}\ast g^{-1}$ is a two-sided inverse of $g\ast h$ in $M$, whence $g\ast h\in M^{\times }$.

Let $g\in M^{\times }$. The equations $g^{-1}\ast g=e=g\ast g^{-1}$ read with $g^{-1}$ as the element being inverted say that $g$ is a two-sided inverse of $g^{-1}$; hence $g^{-1}\in M^{\times }$.

If $M^{\times }=M$ then $M$ is a monoid in which every element is invertible, that is a group; conversely if $M$ is a group then every element of $M$ is invertible, so $M\subseteq M^{\times }$, and $M^{\times }\subseteq M$ always, giving $M^{\times }=M$.

By step 1.2 the set $M^{\times }$ is closed under $\ast$, so $\ast$ restricts to a binary operation on $M^{\times }$, associative because it is associative on $M$.

By step 1.1 the element $e$ lies in $M^{\times }$, and $e\ast x=x=x\ast e$ holds for every $x\in M^{\times }$ because it holds for every $x\in M$; so $(M^{\times },\ast ,e)$ is a monoid.

Every $g\in M^{\times }$ is invertible in $M^{\times }$: its inverse $g^{-1}$ lies in $M^{\times }$ by step 1.3, and the two equations $g^{-1}\ast g=e=g\ast g^{-1}$ are equations between elements of $M^{\times }$. Hence $(M^{\times },\ast ,e)$ is a group.

The units of $M$ form a group under the restricted operation, with the same identity, and this group is all of $M$ exactly when $M$ is a group.