The first method mimics YCor's, but it uses the theory of right-angled Artin groups[right-angled Artin groups][1] (aka partially commutative groups, or graph groups) instead of Bass-Serre theory. The second is prompted by YCor's proof and consists in the computation of the cohomological dimension of $G_n$ which turns out to be $n + 1$. These two methods enable us to retrieve YCor's stronger statement
The last method is the computation of the second integral homology group, also called Schur multiplier[Schur multiplier][2], of $G_n$. We will establish $\text{H}_2(G_n, \mathbb{Z}) \simeq \mathbb{Z}^n$, which yields, as we will see, some other benefit.
Note that the group $S(\Gamma_n)$ is just the kernel $K_n$ of Jim Beck's answer to the initial MSE question.
Proof of Claim 1. Setting as YCor, $s_k \Doteq a^{-k}ba^{k}$ for all $k \in \mathbb{Z}$, we obtain in a similar way $$G_n = \langle a, s_i, \, i \in \mathbb{Z} \,\vert\, a^{-1}s_ka = s_{k + 1},\,[s_i, s_j] = 1 \text{ for all } k, i, j \in \mathbb{Z} \text{ with } \vert i - j \vert \le n \rangle.$$
Hence $G_n \simeq S(\Gamma_n) \rtimes \mathbb{Z}$.
To prove the second part of the claim, we consider the basis $t_1 = \sigma_1a^{i_1}, \dots, t_k = \sigma_k a^{i_k}$, with $\sigma_i \in S(\Gamma_n) $, of a free Abelian subgroup of $G_n$ of rank $k > 1$. Replacing, if needed, $(t_1, \dots, t_k)$ by a Nielsen-equivalent $k$-tuple, we can assume that $i_j = 0$ for every $j > 1$ (use the projection $G_n \twoheadrightarrow \mathbb{Z}$ and the Euclidean algorithm in $\mathbb{Z}$). It follows from the Normal Form Theorem of right-angled Artin group [2] that $t_1$ commutes with $t_2$ only if $i_1 = 0$. Thus $\langle t_1, \dots, t_k \rangle \subset S(\Gamma_n)$. It is well-known that for $S(\Gamma)$, the right-angled Artin group associated to a graph $\Gamma$, the number $a(S(\Gamma))$ is the clique number of $\Gamma$, that is the number of vertices in a maximal complete subgraph of $\Gamma$. Obviously, this number is $n + 1$ for $\Gamma_n$.
Proof. Given a group $G$, a subgroup $S \subset G$ and an injective homomorphism $\tau: S \rightarrow G$, let $T = \tau(S)$ and let $\tilde{G}$ denote the corresponding HNN extension, i.e., $\tilde{G} = \langle G, a \, \vert \, a^{-1}sa = \tau(s), \text{ for all } s \in S \rangle$. By [Theorem 2.12, 1], there is a Mayer-Vietoris sequence
$$
\cdots
\rightarrow
\text{H}^{q - 1}(S, \mathbb{Z})
\mathop{\rightarrow}^{\delta}
\text{H}^q(\tilde{G}, \mathbb{Z})
\mathop{\rightarrow}^{\text{res}}
\text{H}^q(G,\mathbb{Z})
\mathop{\rightarrow}^{\text{res}_S - \tau^{\ast} \circ \text{res}_T} \text{H}^q(S, \mathbb{Z})
\rightarrow
\cdots
$$
Set $G = \langle s_0, \dots, s_n \rangle$, the free Abelian group generated by the $s_i$ for $0 \le i \le n$, and $S = \langle s_0, \dots, s_{n - 1} \rangle$, $\tau: S \rightarrow G$ is the homomorphism induced by the right shift of indices. As shown by YCor, we have then $\tilde{G} = G_n$.
Since $\text{cd}(\mathbb{Z}^q) = q$, inspecting the above exact sequence in dimensions $q = n, n + 1$$q \ge n$ yields the result.
Eventually, we note that the answer to OP's question cannot be too simple as we have $$G_n/G_n' \simeq \mathbb{Z}^2,\quad G_n/G_n'' \simeq \mathbb{Z} \wr \mathbb{Z}$$ for every $n$. To get the second isomorphism, observe that $r_n \equiv r_1 \cdot r_1^a \cdots r_1^{a^{n - 1}} \mod F(a, b)''$$r_n \equiv r_1 \cdot r_1^a \cdots r_1^{a^{n - 1}} \text{mod } (F(a, b))''$.
