User21820 gives an excellent answer, showing that we derive a contradiction so long as we assume {∅} exists. I think that's sufficient to answer your question, and you should accept it if you agree. I felt inclined to give a more in-depth answer, though, since the existence of {∅} is technically not a necessary consequence of your axiom. As I pointed out in a comment, we actually have a trivial model of your axiom provided we assume ∅ is the only set that exists. I'm going to prove that this is the only possible model of your theory. As demonstrated by user21820, we have ∅={{x,y} : ⊥∧¬⊥}, so at least ∅ must exist. In pursuit of contradiction, suppose there exists a nonempty set c....
Given any set n, define S(n)={{x,y} : x=n ∧ y≠n}, and notice that necessarily S(n)≠∅ regardless of n. Indeed, if n=∅ then {n,c}∈S, and otherwise {n,∅}∈S, so in any case we have S(n)≠∅. This also verifies that {a,b}∈S(a) so long as a≠b, hence the unordered pair {a,b} exists so long as a≠b. We also observe that ∅∉S(n), which works since all p∈S(n) obey n∈p and thus p≠∅. Using these facts, we infer the existence of at least three distinct objects: ∅, and S(∅), and finally {∅,S(∅)}.
Since there are at least three distinct objects, we can prove the S function is injective. Indeed, suppose we have S(n)=S(m), then we can find an object k such that both n≠k and m≠k. Since n≠k then {n,k}∈S(n)=S(m) and thus {n,k}∈S(m), and since m≠k then necessarily n=m.
Now, for a proposition P, define the notation [P]={{x,y} : Q(x) & ¬Q(y)}, where Q(x) means ∃z, P(z) & x=S(z). Notice that necessarily ¬Q(∅), and thus we have {S(x),∅}∈[P] if and only if P(x) holds. Finally, we derive our contradiction by letting R=[{S(x),∅}∉x], and asking whether or not {S(R),∅}∈R. By definition of R, we have {S(R),∅}∈R if and only if R satisfies the proposition {S(x),∅}∉x in place of x, and this is an outright contradiction.
The higher-level logic here is actually pretty interesting, and it's how I came up with this proof. All Russel-like paradoxes are ultimately just a variation on Cantor's diagonal argument, which proves |U|<|2^U| for any set U. Some structures are inherently larger (read: more complex) than others, in that they resist compression beyond a certain point. This holds for infinite structures just as well as it does for finite structures.
Let U denote the class of all objects in your domain of discourse, and let 2^U denote the hyperclass which contains all the subclasses of U. Your axiom induces a function from 2^U to U, given by F(C)={{x,y} : x∈C ∧ y∉C}. This function is nearly injective: If we have F(A)=F(B), then either A=B or else A=U\B. If we require that A and B agree on the membership of at least one object, such as requiring that both ∅∉A and ∅∉B, then having F(A)=F(B) simply implies A=B. Because of this, your axiom gives us an injection witnessing 2^|U-1|≤|U|. This is already pretty close to impossible.
What I did was show that |U|≤|U-1|, by constructing a function (namely S) which was injective but not surjective. Incidentally, the function S is literally just S(n)=F({n}), which is injective due to the near injectivity of F (and the fact that 3≤|U|), and fails to be surjective since we never have S(n)=∅. This induces an injection 2^|U| ≤ 2^|U-1|, which composes with F to give an injection 2^|U|≤|U|, which was exactly my [P] notation. From there, I simply deployed Cantor's theorem to derive a contradiction.
