Showing that a subset of a ring with a certain property is a union of prime ideals.

Question

Any subset $S $ of a ring $R$ such that $xy \in S \implies x \in S$ or $y \in S$ is a union of prime ideals of $R$.

Prime ideals of a ring are defined as those $I$ such that $xy \in I \implies x \in I, y \in I$. This property extends easily to unions of prime ideals.

Is the converse true? If $S \subset R$ has the prime property, then is $S$ a union of prime ideals?

In other words,

The set of prime ideals form a basis for open sets in a topology on $R$. Define $S \subset \Bbb{R}$ to be a prime subset if it has the prime property $xy \in S \implies x \in S$ or $y \in S$. Then are prime subsets always the open sets generated by the prime ideals?

Answer 1

No, for example take $R=\Bbb Z$. Then $\Bbb Z\setminus 2\Bbb Z$ has the prime property but there is no ideal which contains $1$ and does not contain $2$.

Answer 2

The smallest counter-example is $R=\Bbb Z/3\Bbb Z$ and $S=\{-1\}$: If $x,y\in\{0,1\}$, then also $xy\in\{0,1\}$.