Even if you don't assume the existence of AC, that doesn't mean the opposite is true. The existence of choice function when one is not constructible, or the existence of a set that is defined as having no choice function, are first-order statements that are independent from the rest of ZF.
If you can construct a theorem about sets without using AC, it can be used both in models that incorporate AC, and those that incorporate its negation. By discussing a zero product of non-zero sets, you already assume that such a construct exists. The property these sets would have, is the property you just mentioned. It's not necessarily intuitive. It doesn't stem from other properties you are familiar with.
Let me give you an example to demonstrate what I'm saying. ℤ can be seen as an extension of ℕ. In the set ℕ, we can say that there exists no number $n$ holding $n < 0$. In the set ℤ there exists such a number.
We could then prove theorems such as, $a + b ≥ a$. In ℤ, where the opposite holds, one can prove that there exist $a,b$ such as that $a+b<a$. These theorems are mutually exclusive given the other laws of ℕ and ℤ. However, you could also imagine a set $ℕ_?$ which would not posit anything at all about the existence of the number $n$ holding $n<0$. While in ℕ and ℤ either the statement or its inverse will hold, $N_?$ doesn't guarantee either.
Technically, you could define $ℕ_?$ as an extension to $ℕ$ in which the number 0 has zero or more predecessors, e.g. $\{-1,-2,-3\}∪ℕ$. It may or may not contain a number $n$ that satisfies $n < 0$, and there is no way of showing that it does, but it is unbounded from above, well-ordered, and has a least element, though there are serious limitations about how it can be used.
A proof in $N_?$ could be that for every number $n$ there exists a number $m$ such as that $m>n$, because that is independent of the existence of negative numbers, only on the fact that $N_?$ is well-ordered and is unbounded from above.
You could thus similarly ask that, given you know only ℕ, what sorts of qualities the numbers $a,b$ would have, if they hold $a+b<a$? The question is problematic. In ℤ we can say that negative numbers have certain definite qualities because that's how we define them. In $ℕ_?$ it's not possible to answer the question, except tautologically, because we don't know if any elements that satisfy the relation exist.
You could give a weaker version of AC in which, say, all countable sets have a choice function. In that case, you would definitely be able to say that if the product of two non-empty sets is empty, then one of them must be uncountable. Otherwise, you really cannot, by definition, derive anything at all about sets that satisfy the property.