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


1. if one has finite index in the other, or
2. 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))?



Quasi-isometric groups without common virtual geometric model
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](http://homepages.math.uic.edu/~furman/preprints/ME.pdf)), 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](https://arxiv.org/pdf/1405.2667.pdf) it is shown that the groups given by ThiKu in his answer are not ME$^p$. Again, it follows that they are not E. 


