If we have a group $G$ defined by: $G=\langle a,b\mid b^2=1\rangle$ then I first need to construct the cayley graph of this, now I think that this is going to look like the "telephone pole" metric space (the cayley graph of the group $F(a,b)$) but vertical it will always stop after one application of $b$
I now want to show that this is quasi-isometric to $\mathbb{Z}$. Can I just do this by mapping each vertex $ab^iab^j\cdots$ to $i+j$ in the graph of $\mathbb{Z}$ and sending $b$ to 0?
The claim is false as written.
$G = \langle a, b \mid b^2 = 1 \rangle \cong \mathbb{Z} \ast \mathbb{Z}_2$. This group is a free product $A \ast_C B$, where $\mid A/C \mid \geq 3$ and $\mid B/C \mid \geq 2$. ($A=\mathbb{Z}, B=\mathbb{Z}_2, C=\{1\}$). By Stalling's Theorem, this group has infinitely many ends. But $\mathbb{Z}$ has 2 ends. Since ends are a quasi-isometry invariant, G cannot be quasi-isometric to $\mathbb{Z}$.
HOWEVER, what you might have meant is that the abelian group $H = \langle a, b \mid [a,b] = b^2 = 1 \rangle \cong \mathbb{Z} \times \mathbb{Z}_2$ is quasi-isometric to $\mathbb{Z}$ as it has a finite index cyclic subgroup. Its Cayley graph with respect to the standard generating set is a ladder and also easy to construct.