In short, we do not need to adopt this as an axiom. But...
If there are sets at all, the axiom of subsets tells us that there is an empty set: If $x$ is a set, then $\{y\in x\mid y\ne y\}$ is a set, and is empty, since there are no elements $y$ of $x$ for which $y\ne y$. The axiom of extensionality then tells us that there is only one such empty set.
So, the issue is whether we can prove that there are any sets. The axiom of infinity tells us that there is a set (which is infinite, or inductive, or whatever formalization you use). But this seems like a terrible overkill to check that there are sets, to postulate that there are infinitely many.
Some people prefer to have an axiom that states that there are sets. Of course, some people then just prefer to have an axiom that states that there is an empty set, so we at once have that there are sets, and avoid having to apply comprehension to check that the empty set exists.
Other people adopt a formalization of first order logic in which we can prove that there are sets. More carefully, most formalizations of logic (certainly the one I prefer) prove as a theorem that the universe of discourse is nonempty. In the context of set theory, this means "there are sets". This is pure logic, before we get to the axioms of set theory. Under this approach, we do not need the axiom that states that there are sets, and the existence of the empty set can be established as explained above.
(The logic proof that there are sets is not particularly illuminating or philosophically significant. Usually, one of the axioms of first order logic is that $\forall x\,(x=x)$. If $\exists x\,(x=x)$ --the formal statement corresponding to "there are sets"-- is false, then $\forall x\,(x\ne x)$. Instantiating, we obtain $x\ne x$, and instantiating the axiom $\forall x\,(x=x)$ we obtain $x=x$, and one of these conclusions is the negation of the other, which is a contradiction. This is not particularly illuminating, because of course we choose our logical axioms and rules of instantiation so that this silly argument can go through, it is not a deep result, and probably we do not gain much insight from it.)
It turns out that yet some others prefer to allow the possibility that there are empty universes of discourse, so their formalization of first order logic is slightly different, and in this case, we have to adopt some axiom to conclude that there is at least one set.
At the end of the day, this is considered a minor matter, more an issue of personal taste than a mathematical question.
As an example, let $X = \{1,2\}$ (where $0=\varnothing, 1= \{0\}, 2=\{0,1\}$). Then
$$X\cap 1 = \varnothing$$
$$X\cap 2 = \{1\}$$
Due to the first equality, our set satisfies the axiom of regularity. $X$ being disjoint with $1$ does not imply that $X$ does not contain $1$; $X\cap 1 = \varnothing$ is a completely different statement from $1\notin X$. Indeed, the former holds, as the only element of $1$ is $0$, which is not an element of $X$.
Best Answer
Yes, $A$ is just the emptyset.
We don't even need to appeal to Foundation to show this: all we need is that the emptyset exists. To be in $A$, you would have to be in every set, so in particular you would have to be in the emptyset - but that's clearly impossible.
Similarly, $B$ is just the emptyset (at least, in ZFC): to be in $B$ is to be a universal set, and (in ZFC) there aren't any of those.
Note that this is a little more finicky than the analysis of $A$: there are set theories which do have a universal set, such as NF, and in such theories the class $B$ is not empty. In all the set theories I know, however, the class $B$ is a set (whether empty or not): in particular, as long as we have $(i)$ Extensionality, $(ii)$ Emptyset, and $(iii)$ Singletons, we're good (if there are no universal sets then $B$ is the empty class, which is a set by $(ii)$; if there is at least one universal set, then there is exactly one universal set by $(i)$ since any two universal sets have the same elements, and so $B$ is the class containing just that universal set, which is a set by $(iii)$), and these are fairly un-controversial axioms.
(OK fine there are some interesting set theories without Extensionality; but still, in all the natural examples I'm aware of $B$ is a set.)
Really, there's a slight abuse going on here: a priori $A$ and $B$ are just classes. What's really going on is that I have the formulas $$\alpha(x)\equiv \forall y(x\in y)\quad\mbox{and}\quad \beta(x)\equiv\forall y(y\in x)$$ defining the classes $A$ and $B$ respectively; I prove in ZFC that "each class is empty," that is, that $$\forall x(\neg\alpha(x))\quad\mbox{and}\quad\forall x(\neg\beta(x)).$$
This now lets me prove "There is a set $U$ such that for all $x$ we have $x\in U\iff \alpha(x)$" - namely, take $U=\emptyset$ - and similarly for $\beta$. This is the "under-the-hood" version of proving that an expression in set-builder notation actually defines a set: we show that there is a set which is co-extensive with the defining formula of the class. In my opinion, this is an example of a situation where set-builder notation being used at the beginning makes things harder to follow: really, we should be asking (in the case of $A$) "Is there a set $U$ such that for all $x$ we have $x\in U\iff \forall y(x\in y)$?" which clearly separates formulas/classes and sets.