Are there products in the category of $\sigma$-algebras and (reversed) $\sigma$-homomorphisms?

The products you're looking for are just coproducts in the category of $\sigma$-algebras. These can be shown to exist by various high-level arguments: for instance, $\sigma$-algebras are models of a generalized algebraic theory, and such categories always admit all limits and colimits.

The general recipe for constructing coproducts of models of an algebraic theory is as follows. For objects $X$ and $Y$ that are free, say $X=F(S)$ and $Y=F(T)$ for sets $S$ and $T$, the coproduct $X\coprod Y$ must be the free algebra $F(S\coprod T)$, since $F$ is a left adjoint and preserves coproducts. If $X$ and $Y$ are general algebras, then we give them a presentation. Categorically, this corresponds to writing $X$ as a (reflexive) coequalizer of free algebras $X''\rightrightarrows X'$, where $X'$ is free on a generating set for $X$ and $X''$ is free on a set of relations. Now we can construct $X\coprod Y$ as the coequalizer of the coproduct of presentations for $X$ and $Y$, that is of the canonical maps $X''\coprod Y''\rightrightarrows X'\coprod Y'$. In short, we construct coproducts by taking the disjoint union of presentations. This is likely familar from the example of the free product of groups.

Your example falls into this description: we give $\mathcal A$ and $\mathcal B$ the maximal presentations with all elements generating and all possible relations listed. Then the coproduct $\mathcal A\coprod \mathcal B$ of $\sigma$-algebras is generated by the disjoint union $\mathcal A\sqcup \mathcal B$ of the underlying sets. This corresponds in your picture to generating $\mathcal A\coprod \mathcal B$ by the elements $A\times Y$ and $X\times B$ instead of $(A,B)$, which works fine. However, I'm not completely sure whether the $\sigma$-algebra you suggest is actually the coproduct of $\sigma$-algebras. I think the analogous question fails for sufficiently unusual topological spaces-the topology on the product of the spaces may not be the coproduct of the topologies. However, such a coproduct certainly does exist. It's just hard work to get an explicit handle on it, which is one key reason to prefer measurable spaces to abstract $\sigma$-algebras.

EDIT: I missed a subtlety resulting from your presentation, which is that it's not immediate that the coproduct of $\sigma$-algebras of sets should coincide with the coproduct of abstract $\sigma$-algebras. Unlike the case for Boolean algebras, not every abstract $\sigma$-algebra can be represented as a $\sigma$-algebra of sets, and so the question is serious. Apparently, every abstract $\sigma$-algebra can instead be represented as the quotient of a canonical $\sigma$-algebra of sets by a $\sigma$-ideal. It seems plausible to me that this quotient is the counit of an adjunction whose left adjoint is the inclusion of $\sigma$-algebras of sets into abstract $\sigma$-algebras. If that is the case, then the construction above gives the correct construction of the coproduct. Unfortunately, the reference I have for this claim is very dated, and I can't easily extract the desired adjunction from it. Here it is: http://www.ams.org/journals/bull/1947-53-08/S0002-9904-1947-08866-2/S0002-9904-1947-08866-2.pdf

In any case, it's unclear to me that there's any reason to think solely about $\sigma$-algebras of sets, if you're only actually using the maps of abstract $\sigma$-algebras. That's another way to avoid this worry, but maybe you have a concrete reason (no pun intended) no to do that.