Here’s a proof that doesn’t involve the quotient $R/A$.
Suppose that $A$ is not prime; then there are $a,b\in R\setminus A$ such that $ab\in A$. Let $B$ be the ideal generated by $A \cup \{a\}$; $B = \{x+ar: x\in A\text{ and }r\in R\}$. Clearly $A \subsetneq B$, so $B = R$, $1_R \in B$, and hence $1_R = x + ar$ for some $x\in A$ and $r\in R$. Then $$b = b1_R = b(x+ar) = bx + bar.$$ But $bx \in bA \subseteq RA = A$, and $bar \in Ar \subseteq AR = A$, so $b \in A$. This contradiction shows that $A$ is prime.
I propose the following: suppose $\,U\,$ is not prime, thus there exist $$\,x,y\in R \text{ such that }x,y\notin U\,,\,xy\in U\,.$$ Define now $B:=U+\langle y\rangle$.
By maximality of $\,U\,$ we have that $\,B\,$ is f.g., say $$\,B=\Bigl\langle u_i+r_iy\,,\,1\leq i\leq k\,,\,k\in\mathbb{N}\,\,;\,\,u_i\in U\,,\,r_i\in R\Bigr\rangle$$ and let now $$U_y:=\{s\in R\,\,;\,\,sy\in U\}.$$ (1) Check that $\,U_y\,$ is a proper ideal in $\,R\,$.
(2) Show that $\,U_y\,$ is f.g.
Put $\,U_y=\langle s_1,\ldots,s_m\rangle\,$, and take $\,u\in U\Longrightarrow\,\exists v_1,\ldots,v_k, t_1,\ldots,t_k\in R\,\,s.t.$$$u=\sum_{n=1}^kv_nu_n+\sum_{n=1}^kt_nr_ny.$$
(3) Show that $\displaystyle{\sum_{n=1}^kt_nr_n}\in U_y.$
(4) Putting $\,\Omega:=\{u_i,\ldots,u_k,ys_1,\ldots,ys_m\}\,$, derive the contradiction $\,U=\langle\Omega\rangle$.
Best Answer
I assume that all multiplicative subsets discussed here are disjoint from $\{0\}$.
Let me write $S$ for the maximal multiplicative subset of $R$.
Step 1: I claim that $S^{-1} R$ is a local ring.
Indeed, if not, there would be at least two maximal ideals in the localization, i.e., two prime ideals $\mathfrak{p}_1 \neq \mathfrak{p}_2$, each maximal with respect to being disjoint from $S$. Thus $S$ is contained in both $R \setminus \mathfrak{p}_1$ and $R \setminus \mathfrak{p}_2$, but these are two different sets so at least one of the containments must be proper, contradicting the maximality of $S$.
Step 2: Therefore there is a unique maximal ideal in the localization, which corresponds to a prime ideal $\mathfrak{p}$ of $R$ which is disjoint from $S$. Arguing as above, maximality of $S$ implies $S = R \setminus \mathfrak{p}$.
Remark 1: When $R$ is a domain, the unique maximal multiplicative subset is obviously $R \setminus \{0\}$. In this case it is tempting to construe "maximal multiplicative subset" to mean "multiplicative subset maximal with respect to being properly contained in $R \setminus \{0\}$."
Remark 2: Conversely, the primes $\mathfrak{p}$ such that $R \setminus \mathfrak{p}$ is maximal are precisely the primes which are minimal with respect to containing $\{0\}$. (When $R$ is a domain and the question is reconstrued as above, we want $\mathfrak{p}$ to be minimal with respect to properly containing $\{0\}$.)
Mariano's answer makes this clear as well.