One can construct examples of such groups via combinatorial group theory. 

For a group $G$, let us construct a group $G^R_i$ so that $G < G^R_i$ and for all $x,y\in G, n\leq i$, there exists $z\in G^R_n$ such that $x^ny^n=z^n$. 

To create such a group, we simply add a new element $g_{x,y,n}$ for each such pair $(x,y)\in G\times G, n\leq i$, so that $(g_{x,y,n})^n=x^ny^n$. If $x^ny^n$ has infinite order in $G$, then we see that this is an amalgamated free product with $\mathbb{Z}$, hence $G$ injects into $\langle G, g_{x,y,n} | (g_{x,y,n})^n=x^ny^n \rangle$. If $x^ny^n$ has order $m$, then we also assume that $g_{x,y,n}$ has order $nm$, and hence we also get a free amalgamated product of $G$ with $\mathbb{Z}/nm\mathbb{Z}$ over $\mathbb{Z}/m\mathbb{Z}$. Taking a union of these amalgamated products, we see that $G$ will inject in the group $$\langle G, g_{x,y,n} | (g_{x,y,n})^n=x^ny^n, x,y\in G\times G, n\leq i \rangle = G^R_i.$$ 

 Now, iterate, taking $G_0=G, G_{i+1}=(G_i)^R_i$, and $\hat{G} = \underset{i}{\cup}\ G_i$. Any $x,y\in \hat{G}$ will lie in $G_i$ for some $i \geq n$, and hence $x^ny^n=z^n$ will have a solution $z\in G_{i+1}$. 

This shows that if you allow (presumably) infinitely generated groups, your class of groups can contain any given group as a subgroup. In particular, your condition does not give a variety of groups (such as the $n$-Abelian groups), although one could ask whether it is a union of varieties (such as groups satifying the identity $x^ny^n = (R(x,y))^n$ for some $R(x,y)\in \langle x,y\rangle$).