Isomorphism does not make an inclusion elementary

The positive tail is nonempty since $1\in A$, and restricting $<$ to $A^2$ makes it a substructure: there are no constant or function symbols requiring further closure. Define $f:\omega \to A$ by $f(n)=n+1$. Each positive natural is uniquely a successor, so $g:A\to \omega$ defined by $g(n+1)=n$ satisfies $g(f(n))=n$ and $f(g(m))=m$. Also $n<k$ iff $n+1<k+1$: adding one preserves strict natural-number order, and if $n\ge k$ then $n+1\ge k+1$. Hence $f$ is a bijection preserving and reflecting the sole relation and is an isomorphism. In particular $f(0)=1$; it is different from inclusion.

At the parameter $1\in A$, the formula $\mathrm{\exists }y(y<x)$ holds in $\mathcal{M}$ because $0<1$. It fails in $\mathcal{A}$: every $y\in A$ is a positive natural, so $1\le y$ and $y<1$ is false. Thus inclusion fails to preserve and reflect this formula's truth and is not elementary, despite the isomorphism constructed in step 1.1.