Rings with rectangles with same perimeter but different surface

Question

Could someone find an example of a commutative unitary ring $(R,+,\cdot,0,1)$, that is not an integral domain, with an element $x\neq 1,0$ such that $$\{a+b\mid a\cdot b=x\,\wedge\,a,b\neq 1\}$$ is a singleton? Or there's not such a ring because of the request hides a contradiction?

Answer 1

Let $R$ be the power set of $\{ 0, 1 \}$, with symmetric difference $\triangle$ and intersection $\cap$ as its addition and multiplication operations, respectively. This is a commutative, unitary ring, with $1_R=\{0,1\}$ and $0_R=\varnothing$. It is not an integral domain since, for instance, $\{ 0 \} \triangle \{ 0 \} = \varnothing$.

Letting $X = \{ 0 \}$, the set in question is $$\{ A \triangle B \mid A \cap B = \{ 0 \} \ \text{and} \ A,B \ne \{ 0, 1 \} \}$$

But there are unique $A,B \in R$ with $A \cap B = \{ 0 \}$ and $A,B \ne \{ 0,1 \}$, namely $A=B=\{0\}$, and so the set in mind is the singleton $\{ \varnothing \}$, since $\{ 0 \} \triangle \{ 0 \} = \varnothing$.