Commensurable, Quasi-Isometric and Finitely Generated Groups Metric Spaces of Non-Positive Curvature,
Book by André Haefliger and Martin Bridson, page $141$.
Two groups are said to be Commensurable if they contain isomorphic subgroups of finite index. Commensurable groups are Quasi-isometric.
The problem is two commensurable groups need not be finitely generated and the concept of quasi-isometric is only for finitely generated groups (Caylay graphs).
 A: There's a generalization to arbitrary groups.
Say that $f:G\to H$ is a coarse map if for every finite subset $F$ of $G$ there exists a finite subset $F'$ of $H$ such that $f(gF)\subset f(g)F'$ for all $x\in G$.
This is stable under composition.
Say that two maps $f,f':G\to H$ are equivalent ($\sim$) if there exists a finite subset $F$ of $H$ such that $f'(g)\in f(g)F$ for all $g\in G$.
Say that $f$ is a coarse equivalence is there exists $f':H\to G$ another coarse map such that $f'\circ f\sim \mathrm{Id}_G$ and $f\circ f'\sim\mathrm{Id}_H$.
Note that these are maps inducing an isomorphism in the category whose objects are groups and arrows are coarse maps modulo $\sim$. In particular, the existence of a coarse equivalence is an equivalence relation between spaces.
Alternatively, a coarse map $f:X\to Y$ is a coarse equivalence if and only if for every finite subset $F$ of $H$ there exists a finite subset $F'$ of $G$ such that for every $g\in G$ we have $f^{-1}(f(g)F')\subset gF$, and there exists a finite subset $F'$ of $H$ such that $f(G)F'=H$.
It is not hard to prove that a coarse equivalence between finitely generated groups is the same as a quasi-isometry. Also, if two groups are coarsely equivalent and one is finitely generated, then so is the other one.
A group homomorphism $G\to H$ is always a coarse map (clear), and is a coarse equivalence iff it has finite kernel and image of finite index. In particular, commensurable (arbitrary) groups are coarsely equivalent.
