Coproduct in an alternative category of groups

This category does not have coproducts. The simplest example is that it has no initial object: an initial object would be a group $G$ such that for any group $H$ there is exactly one $(G,H)$-bimodule (up to isomorphism), but this is impossible since there is always a proper class of different $(G,H)$-bimodules. (Here by $(G,H)$-bimodule of course I mean set with commuting left $G$-action and right $H$-action.)

Or, consider a coproduct of two copies of the trivial group. That would be a group $G$ together with two right $G$-modules $A$ and $B$ such that for any group $H$ with two right $H$-modules $C$ and $D$, there is a unique $(G,H)$-bimodule $X$ such that $A\circ X\cong C$ and $B\circ X\cong D$. But, since $A\circ X$ is a quotient of $A\times X$, it is empty iff either $A$ or $X$ is empty. So for instance, if $C$ is empty and $D$ is nonempty, then we find that $B$ must be empty and $A$ must be nonempty, but then we get a contradiction if we swap the roles of $C$ and $D$. A similar argument (with messier notation) shows that actually no coproducts at all exist besides unary coproducts.

This is a standard category to consider, but as the other answer shows, it has some deficiencies. A standard way to repair them is to consider instead the category whose objects are small categories, where a left module is generalized to a covariant functor into sets and dually.

Talking about isomorphism classes is not ideal; this is really a bicategory, with natural transformations as 2-morphisms. It should be clear how this specializes to the case of groups: bimodule homomorphisms. This bicategory does have colimits in the appropriate sense, which are given by the corresponding colimits of categories. Thus the closest thing to a coproduct of two groups $G,H$ in your category is probably the disjoint union of $G$ and $H$, viewed as one-object categories.