Cantor-Bernstein for quasi-isometric embeddings?

Let $C_n$ be a cyclic groups of order $n$. Then the wreath products $C_2\wr\mathbf{Z}$ and $C_3\wr\mathbf{Z}$ embed QI into each other (for the reverse direction, observe that $C_2\wr\mathbf{Z}$ has a subgroup of index 2 isomorphic to $C_2^2\wr\mathbf{Z}$). But $C_2\wr\mathbf{Z}$ and $C_3\wr\mathbf{Z}$ are not QI, this follows from work of Eskin-Fisher-Whyte ($F\wr\mathbf{Z}$ and $F'\wr\mathbf{Z}$ for $F,F'$ finite groups are QI iff $|F|,|F'|$ have some common power.)

In addition to Yves' answer, I would like to mention right-angled Artin groups as finitely presented and torsion-free counterexamples. I recall that, given a simplicial graph $\Gamma$, the associated right-angled Artin group is defined by the following presentation: $$A_\Gamma = \langle V(\Gamma) \mid [u,v]=1, \ \{ u,v \} \in E(\Gamma) \rangle$$ where $V(\Gamma)$ and $E(\Gamma)$ denote the vertex- and edge-sets of $\Gamma$ respectively.

First, a general statement. A graph $\Gamma$ is join if there exists a partition $V(\Gamma)= A \sqcup B$ such that $A,B$ are both non-empty and such that any vertex of $A$ is adjacent to any vertex of $B$.

Proposition: Let $\Gamma$ be a finite simplicial graph which is not a join and which is not reduced to a single vertex. Then $A_\Gamma$ contains a quasi-isometrically embedded subgroup isomorphic to the free product $A_\Gamma \ast \mathbb{Z}$.

Sketch of proof. Let $\Gamma^e$ denote the extension graph of $\Gamma$, ie., the graph whose vertices are the conjugates $gug^{-1}$, where $g \in A_\Gamma$ and $u \in V(\Gamma)$, and whose edges link two elements when they commute. In Embedability between right-angled Artin groups, Kim and Koberda prove that, if $\Lambda$ is a finite induced subgraph of $\Gamma^e$, then $A_\Lambda$ embeds into $A_\Gamma$. It turns out that $\Gamma^e$ is unbounded since $\Gamma$ is not a join, so that $\Gamma^e$ contains an induced subgraph isomorphic to $\Gamma \sqcup \{ \text{vertex} \}$, hence $A_\Gamma \ast \mathbb{Z} \leq A_\Gamma$.

I don't know if it follows from their argument that the embedding is quasi-isometric, but, in the alternative proof I gave of their theorem in my thesis (see Section 8.5), it is not difficult to show that the embedding I construct is quasi-isometric. $\square$

Corollary: If $\Gamma$ is a finite connected simplicial graph which is not a join, then $A_\Gamma$ and $A_\Gamma \ast \mathbb{Z}$ quasi-isometrically embeds into each other, but they are not quasi-isometric.

The fact that $A_\Gamma$ and $A_\Gamma \ast \mathbb{Z}$ are not quasi-isometric follows from the observation that $A_\Gamma$ is one-ended (as $\Gamma$ is connected and contains at least two vertices).

Now, a concrete example. Consider the right-angled Artin group $$A= \langle a,b,c,d \mid [a,b]=[b,c]=[c,d]=1 \rangle.$$ Of course, $A$ quasi-isometrically embeds into $A \ast \mathbb{Z}$. Conversely, an embedding as mentioned above shows that the subgroup $$\langle a,b,c,d^2,ada^{-1} \rangle= \langle a,b,c,d^2 \rangle \ast \langle ada^{-1} \rangle \simeq A \ast \mathbb{Z}$$ is quasi-isometrically embedded into $A$, which can also be proved directly. However, $A$ and $A \ast \mathbb{Z}$ are not quasi-isometric since $A$ is one-ended.