The identity you're looking for is
$$ \nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B}) $$
so indeed, if the curl of both factors in the cross product vanish, the divergence of the cross product vanishes, too.
I) The main point is that we usually only consider tensor products $V \otimes W$ of vector spaces $V$, $W$ (as opposed to general sets $V$, $W$). But groups (say $G$, $H$) are often not vector spaces. If we only consider tensor products of vector spaces, then the object $G \otimes H$ is nonsense, mathematically speaking.
With further assumptions on the groups $G$ and $H$, it is sometimes possible to define a tensor product $G \otimes H$ of groups, cf. my Phys.SE answer here and links therein.
II) If $V$ and $W$ are two vector spaces, then the tensor product $V \otimes W$ is again a vector space. Also the direct or Cartesian product $V\times W$ of vector spaces is isomorphic to the direct sum $V \oplus W$ of vector spaces, which is again a vector space.
In fact, if $V$ is a representation space for the group $G$, and $W$ is a representation space for the group $H$, then both the tensor product $V\otimes W$ and the direct sum $V\oplus W$ are representation spaces for the Cartesian product group $G\times H$.
(The direct sum representation space $V\oplus W\cong (V\otimes \mathbb{F}) \oplus(\mathbb{F}\otimes W)$ for the Cartesian product group $G\times H$ can be viewed as a direct sum of two $G\times H$ representation spaces, and is hence a composite concept. Recall that any group has a trivial representation.)
This interplay between the tensor product $V\otimes W$ and the Cartesian product $G\times H$ may persuade some authors into using the misleading notation $G\otimes H$ for the Cartesian product $G\times H$. Unfortunately, this often happens in physics and in category theory.
III) In contrast to groups, note that Lie algebras (say $\mathfrak{g}$, $\mathfrak{h}$) are always vector spaces, so tensor products $\mathfrak{g}\otimes\mathfrak{h}$ of Lie algebras do make sense. However due to exponentiation, it is typically the direct sum $\mathfrak{g}\oplus\mathfrak{h}$ of Lie algebras that is relevant. If $\exp:\mathfrak{g}\to G$ and $\exp:\mathfrak{h}\to H$ denote exponential maps, then $\exp:\mathfrak{g}\oplus\mathfrak{h}\to G\times H$.
Best Answer
You're used to the definitions$$U\cdot V=U_iV_i,\,(U\times V)_i:=\epsilon_{ijk}U_jV_k.$$(I've used Einstein notation without worrying about index heights.) Similarly,$$\nabla\cdot V=\partial_iV_i,\,(\nabla\times V)_i:=\epsilon_{ijk}\partial_jV_k.$$Since derivatives don't commute with functions, the consequences are slightly more complicated than for "normal" vectors. For example, in$$\nabla\times(U\times V)=U(\nabla\cdot V)-V(\nabla\cdot U)+\color{red}{(V\cdot\nabla)U-(U\cdot\nabla)V},$$the red terms have no "vanilla" analog. On the other hand, derivatives commute with each other, so e.g. $\nabla\cdot\nabla\times V=0$.