Proof of Uniqueness of the Identity Element and of Inverses in a Group

We begin with (i). Since $(G,\cdot)$ is a group, it is a monoid (Groups and Abelian Groups (i)), so $\cdot$ has a neutral element $e$, the identity of the monoid (Monoids (i)). If $e'$ is also a neutral element of $\cdot$, then $e'=e$, because $\cdot$ has at most one neutral element (A Binary Operation Has at Most One Neutral Element (i)). Thus $\cdot$ has exactly one neutral element, namely the identity $e$.

For (ii), let $g\in G$. Every element of a group is a unit of its monoid (Groups and Abelian Groups (i)), so there is $h\in G$ with

$$gh=hg=e$$

(Units of a Monoid (i)), where $e$ is the identity just obtained. The elements $h\in G$ satisfying these equations are precisely the inverses of the unit $g$ in the sense of Units of a Monoid (i), and a unit of a monoid has exactly one inverse (Units of a Monoid: Unique Inverses, Products, and the Group of Units (i)). Hence there is exactly one such $h$.