By TDLC: I mean totally disconnected locally compact group. Consider the conditions, for $G,H$ discrete groups:
(1) there exists a TDLC group $M$ admitting $G,H$ as subgroups, with a compact open subgroup $K$ such that $K\cap G=K\cap H=\{1\}$ and $GK=HK=M$.
(2) $G$ and $H$ are both cocompact lattices, with the same covolume in some common TDLC group;
(3) $G$ and $H$ are both cocompact lattices in some common TDLC group;
(4) $G$ and $H$ are both cocompact lattices in some common locally compact group;
(5) $G$ and $H$ are quasi-isometric.
Then (1)$\Rightarrow$(2)$\Rightarrow$(3)$\Rightarrow$(4)$\Rightarrow$(5).
(for (1)$\Rightarrow$(2), use the automorphism group of the graph as TDLC group)
For $G,H$ finitely generated, (1) is equivalent to (1') $G,H$ admit actions on some common finite degree connected graph, that are simply transitive on the set of vertices, which, in turn, is equivalent to sharing some Cayley graph (Here Cayley graphs are assumed unlabeled, unoriented, with no multiple edges). For (1)$\Rightarrow$(1'), choose a bi-$K$-invariant symmetric subset $S$ of $G$ such that $K\backslash S/K$ is finite, which generates $G$, and use it to define a $G$-invariant graph (=incidence structure) on $G/K$, namely $gK-hK$ if $h\in gS$.
For finitely generated groups, none of the implications can be reversed.
Counterexample to (2)$\Rightarrow$(1): $\mathbf{Z}$ and $\mathbf{Z}\times\mathbf{Z}/2\mathbf{Z}$.
Counterexample to (3)$\Rightarrow$(2): the trivial group and $\mathbf{Z}/2\mathbf{Z}$ (there are infinite ones too).
Counterexample to (4)$\Rightarrow$(3): $\mathrm{Hei}(A)$ denoting the Heisenberg group over the ring $A$, $\mathrm{Hei}(\mathbf{Z}[\sqrt{2}])$ and $\mathrm{Hei}(\mathbf{Z}[\sqrt{3}])$.
Counterexample to (5)$\Rightarrow$(4): if $\Gamma$ is a cocompact lattice in $\mathrm{Sp}_4(\mathbf{R})$ and $\tilde{\Gamma}$ its inverse image in the universal covering, then $\Gamma\times\mathbf{Z}$ and $\tilde{\Gamma}$ are the examples. (Indeed (4) implies that $G$ has Kazhdan's Property T iff $H$ does, but $\tilde{\Gamma}$ has Kazhdan's Property T while $\Gamma\times\mathbf{Z}$. So they don't satisfy (4), while they are quasi-isometric, since they are commable, i.e. one passes from one to another by a chain of cocompact inclusions in both directions.)
Nevertheless, (non-)quasi-isometry already provides a strong obstruction to (1), and the various quasi-isometry invariants we have could be used to extend this chain of implications.
It is using (1) that I can see easily that $\mathbf{Z}$ and $\mathbf{Z}\times\mathbf{Z}/2\mathbf{Z}$ don't satisfy the criterion.
To directly see that $\mathbf{Z}$ and $D_\infty$ satisfy (1): take $M=D_\infty=\langle a,b\mid a^2=b^2=1\rangle$ as ambient group: it admits, as subgroup of index 2, the subgroups $\langle ab\rangle\simeq\mathbf{Z}$, and the subgroup $\langle a,bab\rangle\simeq D_\infty$, and the subgroup $\langle b\rangle$ then plays the role of $K$.
Note that in general, if $G$ and $H$ are subgroup of the same finite index in a common (finitely generated) group $M$, then they satisfy (2); however they may fail to satisfy (1).
Being quasi-isometric (or commable) is transitive. However, conditions from (1) to (4) are not transitive a priori, although I don't have a counterexample in mind for all of them at the moment. So it means that a priori it makes sense to make the same discussion for $n$ given groups.
At least (3) and (4) are not transitive, based on Proposition 19.83 in ArXiv/1212.2229: namely, take $p,q,p',q'$ be odd primes with $\{p,q\}\neq\{p',q'\}$. Let $G$ and $H$ be the (virtually free) free products $C_p\ast C_q$ and $C_{p'}\ast C_{q'}$. Then the pair $(G,H)$ fails to satisfy (4) (and hence fail (3)), by that proposition. However, if $n$ is large enough, the free group $F_{1+n!}$ is isomorphic to a finite index subgroup of both $G$ and $H$, and hence the pairs $(G,F_{1+n!})$ and $(F_{1+n!},H)$ both satisfy (3) and (4).
Best Answer
Here the short proof of the fact that every f.g. group of subexponential growth is amenable.
Let $S$ be a symmetric generating subset with 1. If $G$ has subexponential growth, then clearly $\liminf |S^{n+1}|/|S^{n}|=1$. So we can extract from $(S^n)$ a Følner sequence.
The converse fails: many f.g. solvable groups have exponential growth.