TITLE: Quasi-isometric groups without common virtual geometric model
QUESTION [11 upvotes]: Are there known examples of finitely generated groups $G$ and $H$ that are quasi-isometric but do not admit finite-index subgroups $G'<G$ and $H'<H$ such that both $G'$ and $H'$ admit proper and cocompact actions on the same geodesic metric space $X$?
For context: it is known and not hard to see that groups are quasi-isometric

if one has finite index in the other, or
if both act properly and cocompactly on the same geodesic metric space.

and (1) is really a special case of (2).
So one could say that $G$ and $H$ are elementarily quasi-isometric if they admit $G'<G$ and $H'<H$ as above and ask whether any two quasi-isometric groups are elementarily quasi-isometric. Surely the answer is "no" and the initial question is about known counterexamples.
Additional challenges:

Can the groups be taken to be of type $F$ (or, say,  finitely presented, torsion-free)?
Being "elementarily quasi-isometric" is not (obviously) a transitive relation. Can the groups be taken to not be related even by the transitive closure (= the transitive closure of (2))?

REPLY [7 votes]: The question was already answered by ThiKu and YCor.
I am adding this just to make a conceptual remark.
The OP asks about groups which are QI, but not in an "elementary fashion".
As YCor pointed out in his answer, if $G$ acts properly and cocompactly on a geodesic metric space $X$ then it could be seen as a cocompact lattice in the (locally compact group) of isometries of $X$. 
It thus makes sense to extend our notion of "elementary-ness" as follows:
Definition: Two groups are elementary equivalent if they are in the relation (E) which is the transitive closure of the relation "being cocompact lattices in the same locally compact group".
Finding groups which are QI but not E will answer the OP question. Still, the relation E is defined in a non explicit way. It makes sense to extend it. That is, to find a courser equivalence relation that contains it, and is easier to handle. A good example is the following:
Definition: Two groups are Measure Equivalent (ME) if they act on an infinite measure space, preserving the measure, the actions commute and each has a finite fundamental domain.
The exact relationship between QI and ME is tricky. Both contain the relation E defined above, but none include the other.
However, it is not hard to see that property (T) is an ME invariant (this is Corollary 1.4 in Furman's paper), while it is possible for a (T) group to be QI to a non-(T) group. 
For an example, see (1) in YCor's answer.
This example thus addresses all OP's challenges.
The relation ME admits various refinements called ME$^p$, where $1\leq p\leq \infty$ refers to some inegrability property of an associated function (the norm of the rearrangement cocycle). Again, these are all natural equivalence relation containing E. In Theorem 1.1 of Das-Tessera's paper it is shown that the groups given by ThiKu in his answer are not ME$^p$. Again, it follows that they are not E.
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