Symmetric Difference with Intersection forms Ring

Theorem
Let $S$ be a set.

Then $\left({\mathcal P \left({S}\right), *, \cap}\right)$ is a commutative ring with unity, in which the unity is $S$.

This ring is not an integral domain.

Comment
The same does not apply to symmetric difference and union unless $S = \varnothing$.

For a start, the identity for union and symmetric difference is $\varnothing$ for both, and the only way the identity of both operations in a ring can be the same is if the ring is null.

Unless $S \ne \varnothing$, then this can not be the case.

Also note that union is not distributive over symmetric difference. From Symmetric Difference of Unions, $\left({R \cup T}\right) * \left({S \cup T}\right) = \left({R * S}\right) \setminus T$.