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Under the unital ring convention used elsewhere in this wiki, no such pair exists. In a two-element ring the elements are $0$ and $1$. Additive cancellation forces $1+1=0$, and the identity axiom forces $1\cdot 1=1$. The distributive property forces products involving zero to vanish. Every operation is therefore determined, giving the prime field ${\mathbb{F}}_2$ as the unique possibility up to ring isomorphism.

If the word ring allows a nonunital ring, there are instead two non-isomorphic examples: ${\mathbb{F}}_2$ and the zero-multiplication rng on the additive group $\mathbb{Z}/2\mathbb{Z}$. In the second example the unique nonzero element $u$ satisfies $u+u=0$ and $u^2=0$; every product is zero, so associativity and both distributive laws hold. An isomorphism would have to send $1$ to $u$ because it preserves zero and is bijective, but then it would send $1^2=1$ to both $u^2=0$ and $u$, which is impossible. These exhaust the rngs with two elements, since $u^2$ can only be $0$ or $u$. Specifying the identity convention is essential for this part.

Up to an operation-preserving bijection, there are two rngs with two elements. Write the elements as $0,u$. The additive abelian group must satisfy $u+u=0$. The distributive property forces all products involving zero to vanish, leaving only the choice $u^2=0$ or $u^2=u$. The first gives a zero-multiplication rng; the second gives the prime field ${\mathbb{F}}_2$, with identity $u$. Both satisfy the axioms, and they are not isomorphic because an isomorphism must send the unique nonzero element to the unique nonzero element and preserve its square. In particular, there is just one unital ring with two elements.