The zero element is the empty set and each element is its own additive inverse.
The set S itself is the multiplicative identity and is the only element with a multiplicative inverse.
With A + B = A ∪ B we do not get a ring. The only thing that goes wrong is additive inverses.
The only thing that stops (R3, +, ×) being a ring is that × is not associative.
To see this observe that (u × v) × w is perpendicular to u × v and so in the plane spanned by u and v, while u × (v × w) is perpendicular to v × w and so in the plane spanned by v and w. It is then easy to find a counterexample.
Check it directly [using the fact that if p(1) = q(1) = 0 then (p - q)(1) = 0 and p(1)r(1) = 0 for any polynomial r] or (cleverer!) observe that this is the ideal < x - 1 > generated by the polynomial x - 1.
As usual, this is a commutative ring with identity and so we need only find multiplicative inverses.
1-1 = 1 and 2-1 = 2 since -1 = 2 the "rule" gives x(x + 2) = 1 and we have inverses for two more elements. Squaring this gives (x + 1)(2x + 2) = 1 and so we have two more. Also -x(-x - 2) = 1 and this gives 2x(2x + 1) = 1 and this is the last pair.
A cleverer way to work is to observe that powers of x generate the eight non-zero elements of the ring and then (xm)-1 = x8-m.
The order of any non-zero element is then the same as the order of 1.
In fact, if the polynomial p(x) cannot be factorised ("is irreducible") then using the "rule" p(x) = 0 will give a field.
If the polynomial x2 - x - 1 ∈ Z3[x] could be factored it would have to have linear factors and hence would have a root. But it is easy to check that substituting x = 0, 1, 2 will not give 0.