Well, one way to think about it is this: instead of thinking about matrices, you should think about linear transformations. (The nomenclature below maybe a bit cukoo, since I am not really an algebraist...)
Let $V, W$ be two three-dimensional real (doesn't matter, any field will do; I choose the reals for simplicity of notation) vector spaces. Let $A:V\to W$ and $B:V\to W$ be linear transformations. Treat $V$ to be part of the (graded) tensor product algebra, that is, consider the (infinite) set of vector spaces $\{ \mathbb{R} = \otimes^0 V, V, V\otimes V, V\otimes V\otimes V, \ldots \}$, with the multiplication operation given by the tensor product. Treat $W$ as part of the (graded) exterior algebra, that is the set $\{ \mathbb{R}, W, W\wedge W, W\wedge W\wedge W\}$ (since $W$ has dimension 3, I stop at the third order exterior product; you can generalize this to dimension $n$ if you want) with the multiplication map given by the exterior product.
Your $A\times B$ is the map $V\otimes V \to W\wedge W$ induced by $A,B$ via the tensor product over $V$ and exterior product over $W$. In other words, your $\times$ is the induced bilinear map from $L(V,W)\times L(V,W) \to L(V\otimes V,W\wedge W)$ via the $(V,\otimes)$ and $(W,\wedge)$.
Seen this way it is pretty clear that there should not be, in general, a nice interpretation of $A \times (BC)$ in terms of $A\times B$ and $A\times C$. In general you have here three vector spaces $U,V,W$ with $A:U\to W$, $B:V\to W$ and $C:U\to V$, so that ($BC:U\to W$ and the "cross prodcut" against $A$ can be defined). But then since $B$ and $C$ each has domain/codomain different from $A$, you cannot "factor".
(You ask, but aren't all three-dimensional real vector spaces isomorphic? Yes, but since in the domain side we embed into the tensor algebra and on the codomain side we embed into the exterior algebra, you should think of them as distinct objects.)
I'm sure someone better versed in Category Theory can come along and give a cleaner presentation of this.