If $\mathrm{ord}(g)=n$ then $g^k=e$ iff $k$ is an integer multiple of $n$, the powers $g^0,\dots ,g^{n-1}$ are distinct, and $⟨g⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^j=g^k$ only for $j=k$

$e^x=e$ for every $x\in \mathbb{Z}$. For natural exponents the set of $t$ with $e^t=e$ contains $0$, since $e^0=e$, and is closed under $\sigma$, since $e^{\sigma (t)}=e^{t}e=ee=e$; induction gives it for all naturals. For $x<0$ write $-x=\iota (k)$; then $e^x=(e^k{)}^{-1}=e^{-1}=e$.

Assume $\mathrm{ord}(g)=n$ with $n\ge 1$. Then $n\in S_g$, so $g^n=e$, and no natural $s$ with $1\le s<n$ satisfies $g^s=e$, since $n$ is the least element of $S_g$. Also $\iota (n)>0$, because $n\ge 1$ and $\iota$ preserves the order.

Infinite order. Assume $S_g=\mathrm{\varnothing }$ and suppose $g^j=g^k$ with $j,k\in \mathbb{Z}$ and $j\mathrm{\ne }k$. Put $x:=j-k$, so $x\mathrm{\ne }0$ and $g^x=g^j{g}^{-k}=g^j(g^k{)}^{-1}=e$. Also $g^{-x}=(g^x{)}^{-1}=e^{-1}=e$. By totality one of $x$ and $-x$ is positive; call it $y$, so $g^y=e$ and $y=\iota (s)$ with $s\in \mathbb{N}$ and $s\mathrm{\ne }0$, hence $s\ge 1$. Then $s\in S_g$, contradicting $S_g=\mathrm{\varnothing }$.

The "if" half of claim 1: if $k=q\iota (n)$ for some $q\in \mathbb{Z}$, then $g^k=g^{\iota (n)q}=(g^{\iota (n)}{)}^q=(g^n{)}^q=e^q=e$.

The "only if" half. Suppose $g^k=e$. Divide: $k=q\iota (n)+r$ with $0\le r<\iota (n)$, legitimate since $\iota (n)>0$. Then $e=g^k=g^{q\iota (n)}{g}^r=(g^n{)}^q{g}^r=e^q{g}^r=g^r$. Since $0\le r$, we have $r=\iota (s)$ for a unique $s\in \mathbb{N}$, and $\iota (s)<\iota (n)$ forces $s<n$, because otherwise $n\le s$ and $\iota$ would give $\iota (n)\le \iota (s)$, contradicting antisymmetry. So $g^s=e$ with $s<n$.

Claim 2. Let $i,j\in \mathbb{N}$ with $i<n$, $j<n$ and $g^i=g^j$. By trichotomy we may assume $i\le j$, interchanging the names if necessary, and then $i+t=j$ for some $t\in \mathbb{N}$. Now $g^i{g}^t=g^{i+t}=g^j=g^i=g^{i}e$, so $g^t=e$ by cancellation. Moreover $t\le j$, since $t+i=j$, and $j<n$, so $t<n$. If $t\ge 1$ then $t\in S_g$ with $t<n$, impossible; so $t=0$ and $i=j$.

Therefore in the infinite-order case $g^j=g^k$ forces $j=k$. Moreover $⟨g⟩$ is then not finite: a bijection $\psi :⟨g⟩\to m$ with $m\in \mathbb{N}$ would make $k\mapsto \psi (g^{\iota (k)})$ a map $\sigma (m)\to m$, and that map is injective, since $\iota$ is injective, distinct integer exponents give distinct powers by step 1.3, and $\psi$ is injective; but claim 1 of the pigeonhole principle forbids an injection $\sigma (m)\to m$.

In step 2.2 the case $s\ge 1$ is impossible, since it would put $s$ in $S_g$ below its least element; hence $s=0$, so $r=0$ and $k=q\iota (n)$. With step 2.1 this is claim 1.

Every integer power of $g$ is one of $g^0,\dots ,g^{n-1}$: given $x\in \mathbb{Z}$, divide $x=q\iota (n)+r$ with $0\le r<\iota (n)$, write $r=\iota (s)$ with $s\in \mathbb{N}$ and $s<n$ as in step 2.2, and compute $g^x=(g^n{)}^q{g}^r=e^q{g}^s=g^s$.

Claim 3. By [L2] and step 3.2, $⟨g⟩=\{g^s:s\in \mathbb{N},s<n\}$. The map $\phi :n\to ⟨g⟩$ with $\phi (s)=g^s$ is well defined, the elements of the natural number $n$ being exactly the naturals $s<n$; it is surjective by the displayed description and injective by step 2.3. So $\phi$ is a bijection, $⟨g⟩\approx n$, and $⟨g⟩$ is finite with $\mathrm{\mid }⟨g⟩\mathrm{\mid }=n=\mathrm{ord}(g)$, the value $\mathrm{\mid }⟨g⟩\mathrm{\mid }$ being the unique natural equinumerous with $⟨g⟩$.

Claims 1, 2 and 3 are steps 3.1, 2.3 and 4.1, and the infinite-order statement is steps 1.3 and 2.4.