We consider the product of free groups $$\mathbb{F}_2 \times \mathbb{F}_2 = \langle a,b,c,d \mid [a,b]=[b,c]=[c,d]=[d,a]=1 \rangle.$$ Given some elements $g_1,\ldots,g_n \in \mathbb{F}_2 \times \mathbb{F}_2$, viewed as words over $\{a,b,c,d \}^{\pm 1}$, is there an algorithm to determine if the subgroup $\langle g_1, \ldots,g_n \rangle$ is undistorted (ie. the inclusion induces a quasi-isometric embedding)?
For instance, for an explicit example, is the subgroup $\langle a,b,cd \rangle$ undistorted?
Let $P_1$ and $P_2$ be the projections of $H$ on both factors, and $I_1,I_2$ the intersections; then $P_1/I_1\simeq P_2/I_2$, call this group $Q$. Then
(it's due to Olshanskii and Sapir, see the discussion here). Since word-hyperbolicity is not decidable for finite presentations of 2-generator groups, it follows that the problem whether a subgroup given by inputting finitely generators is undistorted is undecidable. Precisely, if $P=\langle x,y\mid w_i(x,y):i\in I\rangle$ is a finite presentation defining a group $\Gamma_P$, consider the subgroup $H_P$ of $F_2\times F_2$ (where $F_2=\langle x,y\rangle$) generated by $(x,x)$, $(y,y)$, and $(w_i(x,y),1)$, $i\in I$. Then $H_P$ is undistorted iff $\Gamma_P$ is word-hyperbolic.
However it is quite a practical criterion. In the example you provide, using my notation you have the subgroup generated by $(x,1)$, $(1,x)$, $(y,y)$; it corresponds to the presentation $\langle x,y\mid x\rangle$ of the infinite cyclic group, which is definitely word-hyperbolic, and this is consistent with the fact you can check that your group is undistorted.