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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.

Let $e_j$ be the sequence with $1$ in coordinate $j$ and zero elsewhere. In the direct product of rings $R=k^{\mathbb{N}}$, the idealsform a strict ascending chain: every element of $(e_1,\dots ,e_n)$ vanishes after coordinate $n$, whereas $e_{n+1}$ does not. Hence $R$ is not a Noetherian ring.

Nevertheless, each $a=(a_j)\in R$ has a coordinatewise generalized inverse $b=(b_j)$, defined by $b_j=a_j^{-1}$ when $a_j\ne 0$ and $b_j=0$ otherwise. It satisfiesThis is the defining property of a Von Neumann regular ring. Fix any prime ideal $P$, and take $a\in P$. Since $ab\in P$, $1-ab\notin P$; butConsequently $a/1=0$ in the localization at a prime ideal $R_P$. All elements of $P{R}_P$ therefore vanish. By the local ring description in part (a), $P{R}_P$ is the unique maximal ideal, so $R_P$ is a nonzero field. A field has only the ideals zero and itself, and thus is Noetherian.

We have proved the stronger conclusionThe argument applies to every prime ideal, without assuming that it comes from a coordinate projection.

A ring is von Neumann regular if every element $a$ has an element $b$ with $a=aba$. In a commutative ring, this is $a=a^{2}b$. An arbitrary direct product of rings whose factors are fields has this property by taking inverses in each nonzero coordinate. Every localization at a prime ideal of a commutative von Neumann regular ring is a field.