$(\mathbb{Z},\cdot ,1)$ is a commutative monoid whose group of units is $\{1,-1\}$; equivalently $u\mid 1$ holds exactly for $u=1$ and $u=-1$

Multiplication on $\mathbb{Z}$ is a function $\mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}$, hence a binary operation, and it is associative and commutative; $1$ is a two-sided identity, since $x\cdot 1=x$ and, by commutativity, $1\cdot x=x$. So $(\mathbb{Z},\cdot ,1)$ is a commutative monoid.

For $u\in \mathbb{Z}$, being a unit of this monoid means $uv=1$ for some $v$, the two equations $uv=1$ and $vu=1$ being the same by commutativity; and $uv=1$ for some $v$ is precisely $u\mid 1$. So ${\mathbb{Z}}^{\times }=\{u:u\mid 1\}$.

Both $1$ and $-1$ lie in ${\mathbb{Z}}^{\times }$: $1\cdot 1=1$ and $(-1)(-1)=1\cdot 1=1$.

$0\le 1$, since $1=\iota (1)$ lies in the image of $\iota$, which is the set of nonnegative integers; hence $\mathrm{\mid }1\mathrm{\mid }=1$. Also $1\mathrm{\ne }0$, since $\iota$ is injective and $1\mathrm{\ne }0$ in $\mathbb{N}$.

Discreteness of $\mathbb{Z}$: if $0<x$ then $1\le x$. Indeed $x\ge 0$, so $x=\iota (k)$ for some $k\in \mathbb{N}$; $k\mathrm{\ne }0$ because $\iota (0)=0\mathrm{\ne }x$; hence $0<k$ in $\mathbb{N}$, so $1=\sigma (0)\le k$, and applying $\iota$, which preserves the order, gives $1=\iota (1)\le \iota (k)=x$.

$1\mathrm{\ne }-1$: otherwise $1+1=1+(-1)=0$, whereas $1+1=\iota (1)+\iota (1)=\iota (1+1)=\iota (2)$ and $\iota (2)\mathrm{\ne }\iota (0)=0$ because $\iota$ is injective and $2\mathrm{\ne }0$ in $\mathbb{N}$.

Let $u\mid 1$. Since $1\mathrm{\ne }0$, [L6] gives $u\mathrm{\ne }0$ and $\mathrm{\mid }u\mathrm{\mid }\le \mathrm{\mid }1\mathrm{\mid }=1$.

Also $\mathrm{\mid }u\mathrm{\mid }\ge 0$ and $\mathrm{\mid }u\mathrm{\mid }\mathrm{\ne }0$, so $0<\mathrm{\mid }u\mathrm{\mid }$, whence $1\le \mathrm{\mid }u\mathrm{\mid }$; with step 2.1 and antisymmetry this gives $\mathrm{\mid }u\mathrm{\mid }=1$.

From $\mathrm{\mid }u\mathrm{\mid }=1$: if $u\ge 0$ then $u=\mathrm{\mid }u\mathrm{\mid }=1$, and if $u\le 0$ then $-u=\mathrm{\mid }u\mathrm{\mid }=1$, so $u=-(-u)=-1$. By totality one of the two holds, so $u=1$ or $u=-1$.

Combining, ${\mathbb{Z}}^{\times }=\{u:u\mid 1\}=\{1,-1\}$, a two-element set, and by [L4] it is a group under multiplication, with identity $1$ and with each of its elements its own inverse.