Here are two examples of different nature:

(1) let $G$ be a simple Lie group with infinite fundamental group and Property T, e.g., $\mathrm{Sp}_{2n}(\mathbf{R})$ for $n\ge 2$. Let $\Gamma$ be a cocompact lattice in $G$. Let $\tilde{\Gamma}$ its inverse image in the universal covering $\tilde{G}$ of $G$. Then $\tilde{\Gamma}$ and $\Gamma\times\mathbf{Z}$ are quasi-isometric but don't act geometrically (=properly cocompactly) on the same proper metric space.

(1') The conclusion of (1) holds without assuming that $G$ has Property T, covering the case $G=\mathrm{SL}_2(\mathbf{R})$, but the proof is a little longer.

(2) Let $H$ be a simple Lie group, not compact and not locally isomorphic to $\mathrm{SL}_2(\mathbf{R})$. Let $\Gamma_i$, $i=1,2$ be cocompact lattices in $G$ that are not abstractly commensurable (these exist precisely because we have excluded $\mathrm{SL}_2(\mathbf{R})$. Then $\Gamma_1\ast\mathbf{Z}$ and $\Gamma_2\ast\mathbf{Z}$ are quasi-isometric but don't act geometrically (=properly cocompactly) on the same proper metric space.

The examples in (1) and (2) are quite different because in (1) they are commable but not in (2). Morally "commable" is the transitive closure of "acting geometrically on the same space"; see precise definitions below.

Proofs:

(1) If discrete groups $\Gamma$, $\Lambda$ act geometrically on the same proper metric space $X$, say with finite kernels $K,L$, then $\Gamma/K$ and $\Lambda/L$ are cocompact lattices in $\mathrm{Isom}(X)$. Since (Kazhdan's) Property T is invariant (= stable in both directions) under taking quotients by finite normal subgroups and passing from a lattice to a locally compact group, we deduce that $\Gamma$ has Property T iff $\Lambda$ has Property T.

Now in this case, $\tilde{\Gamma}$ is known to have Property T (see Bekka-Harpe-Valette's book), while $\Gamma\times\mathbf{Z}$ doesn't.

On the other hand (in the context of (1) or (1')), they are quasi-isometric because they are **commable**. Recall a copci homomorphism $G\to H$ between compactly generated locally compact (CGLC) groups means a **co**ntinuous **p**roper homomorphism with **c**ocompact **i**mage. Then every copci homomorphism is a quasi-isometry.

Two CGLC groups $G,H$ are **commable** if there exists a finite sequence (called **commability**) of copci homomorphisms (in both directions):
$$G\to G_1\leftarrow G_2\to G_3\leftarrow\dots\to H$$
This implies that $G$ and $H$ are quasi-isometric.

In the setting of (1), let $T$ be a simply connected solvable cocompact subgroup in $G$; the its inverse image in $\tilde{G}$ is a direct product $T\times\mathbf{Z}$. Then we get the commability
$$\tilde{\Gamma}\to\tilde{G}\leftarrow T\times\mathbf{Z}\to G\times\mathbf{Z}\leftarrow\Gamma\times\mathbf{Z}.$$
Hence $\Gamma\times\mathbf{Z}$ and $\tilde{\Gamma}$ are commable, hence quasi-isometric (this is classical: this shows failure of QI-invariance of Property T).

(2) Since $\Gamma_1,\Gamma_2$ are quasi-isometric and non-amenable, they are bilipschitz by a result of Whyte. Hence $\Gamma_1$ and $\Gamma_2$ are bilipschitz, hence quasi-isometric.

That $\Gamma_1$ and $\Gamma_2$ are not commable is an observation of Carette and Tessera, see Section 5B here (arXiv link).

(1') For these additional examples (which include those suggested by ThiKu) it remains to prove that $\tilde{\Gamma}$ and $\Gamma\times\mathbf{Z}$ have no geometric action on a common proper metric space, without assuming that $G$ has Property T. It's not immediate; I'll write the proof later.