Yes. For a group $G$, define $a(G)$ as the greatest $k$ such that $\mathbf{Z}^k$ embeds as a subgroup of $G$.
Then $a(G_n)=n+1$ for all $n\ge 0$. Thus the $G_n$, for $n\ge 0$, are pairwise non-isomorphic.
We have $a(G_n)\ge n+1$ since $s_k=a^{-k}ba^k$, $0\le k\le n$ generate a free abelian subgroup on $n+1$ generators. Let us show the other inequality.
We can add the $s_k$ among the generators, so a presentation is given by $$\langle a,s_0,\dots,s_n\mid a^{-1}s_ia=s_{i+1}:0\le i\le n-1,\;[s_i,s_j]=1,0\le i,j\le n\rangle.$$ This is an HNN presentation, namely the HNN extension of $\mathbf{Z}^{n+1}=\langle s_0,\dots,s_n\rangle$ for the isomorphism between two copies of $\mathbf{Z}^n$ given by $s_i\mapsto s_{i+1}$.
Let $A$ be an abelian subgroup in $G_n$ isomorphic to $\mathbf{Z}^d$. If $A$ has a loxodromic element (for the action on the Bass-Serre tree), then its axis is $A$-invariant and hence a copy of $\mathbf{Z}^{d-1}$ fixes this axis pointwise. In particular, this $\mathbf{Z}^{d-1}$ fixes an edge and hence is conjugate into the edge group, namely $\mathbf{Z}^n$. Hence $d-1\le n$, that is, $d\le n+1$. Otherwise, every element of $A$ fixes a vertex, and since $A$ is finitely generated, this implies that $A$ fixes a vertex. Hence $A$ is conjugate into a vertex group which is isomorphic to $\mathbf{Z}^{n+1}$, hence $d\le n+1$ again.
PS $G_\infty$ is the wreath product $\mathbf{Z}\wr\mathbf{Z}$: this is a motivation for studying these "approximating" groups. In the original MathSE question (linked by the OP), it is checked that the canonical quotient map (the homomorphism given by $a\mapsto a$, $b\mapsto b$) from $G_m\to G_n$ when $m<n$ is not an isomorphism.