gr.group theory - Cayley graphs of finitely generated groups

The answer is no, as expected. The following proof is "joint work" with L. Scheele.
Consider $G=mathbb{Z}$, $K=D_infty$ and $H:=mathbb{Z}times mathbb{Z}/2mathbb{Z}$. Then $G approx K$ and $Kapprox H$, but $G notapprox H.$

Indeed, the Cayley graph associated to ${-1,1}$ for G and the Cayley graph associated to ${s,t}$ where $D_infty=langle s,t: s^2=t^2=1 rangle$ are clearly isometric.

Similarly, the Cayley graph associated to ${s,st,ts}$ for $K$ and the graph associated to ${(0,1),(-1,0),(1,0)}$ for $H$ are isometric.

However, let $S$ be some symmetric generating set for $G$. Then $S_k$, the set of vertices which have distance precisely $kgeq 1$ from the identity, has even cardinality because $S_k$ is invariant under the mapping $x mapsto -x$ and doesn't contain 0.

Now let $T$ be some symmetric generating set for $H$. Let $k_0$ be the distance of $(0,1)$ from the identity in the associated graph. Then $T_{k_0}$, the set of vertices which have distance precisely $k_0$ from the identity, has odd cardinality. Indeed, it is invariant under the mapping $(x,y) mapsto (-x,-y)$ since $T$ is assumed to be symmetric. But $(0,1)$ is a fixed point.