A countable nonstandard model has an element above all numerals

Add one fresh constant $c$ and put $T^{\ast }=\mathrm{Th}(\mathcal{N})\cup \{\bar{n}<c:n<\omega \}$. For a finite subset, let $m$ be the largest index in its inequalities if any occur, and interpret $c$ by $m+1$ in $\mathcal{N}$. Then each required inequality is $n<m+1$, true since $n\le m$; every included sentence of the complete theory is true by definition. If no inequality occurs, interpret $c$ by $0$. Thus every finite subset has a model in the countable expanded signature.

F2 gives an at most countable $N\models T^{\ast }$. Let $M$ be its original-language reduct and $b=c^N$. Reducts keep all interpretations of the old symbols, so $M\models \mathrm{Th}(\mathcal{N})$ and $M\models \bar{n}<b$ for every $n$.

If $h:\mathcal{N}\to M$ were an isomorphism, preservation of $0$ gives $h(0)=0^M$, and preservation of $S$ inductively gives $h(n)={\bar{n}}^M$. Surjectivity would give $b=h(k)={\bar{k}}^M$ for some $k$. But step 2.1 would then say $b<b$, contradicting $M\models \mathrm{\forall }x\mathrm{\neg }(x<x)$, a sentence true in $\mathcal{N}$ and therefore in its theory. Hence no isomorphism exists.