 Solution

Yes: $R$ can have nonzero zero divisors. Let $R=k\times k$, a direct product of rings, where $k$ is any field. The nonzero idempotents $e_1=(1,0)$ and $e_2=(0,1)$ satisfy $e_1{e}_2=0$, so both are zero divisors.

There are exactly two prime ideals:Indeed, any prime ideal must contain $e_1$ or $e_2$, since their product is zero. If it contains $e_1$, it contains $k\times 0$ and corresponds to a prime ideal of the quotient ring $k$, whose only prime ideal is zero; hence it equals $P_2$. The other case gives $P_1$.

For the localization at a prime ideal $P_1$, every denominator $(u,v)\notin P_1$ has $u\ne 0$, andis an isomorphism. It is surjective using constant first-coordinate fractions. It is injective because a numerator with first coordinate zero is annihilated by $e_1\notin P_1$, so its fraction is zero. Interchanging the coordinates proves $R_{P_2}\cong k$.

Thus the localization at a prime ideal is a field in every case, although $R$ has zero divisors. The obstruction to using the previous part's argument is that the two nonzero factors of a zero product can survive at different prime ideals.