Lifting an equation from the localization by clearing denominators (Atiyah-Macdonald 5.12)

In proposition 5.12, Atiyah & Macdonald prove that localization commutes with taking the integral closure. That is, they prove the following:

Let $A \leq B$ be commutative rings, Let $C$ be the integral closure of $A$ in $B$, and let $S$ be a submonoid of $A$. Then $S^{-1}(C)$ is the integral closure of $S^{-1}A$ in $S^{-1}B$.

In the proof, they show that every element $\frac bs \in S^{-1}B$ is integral over $S^{-1}A$, by multiplying the equation $$(\frac bs)^n+\frac {a_1} {s_1}(\frac bs)^{n-1}+...+\frac {a_n}{s_n}=0$$ by $(s \cdot \prod s_i)^n$, and claiming that this gives an integral equation in $A$ for $bs_1...s_n$.

How do we justify going from an equation in the localization to an equation in the original ring, given that $A$ is not necessarily a domain?

After all, the equation in the localization means a "pretty complicated" thing: that the numerator resulting from the common denominator of the expression is annihilated by some element of $S$. Do I have to write explicitly the numerator to make this implication? how is it formally justified?

From $$\left(\frac bs\right)^n+\frac{a_1} {s_1}\left(\frac bs\right)^{n-1}+\cdots+\frac {a_n}{s_n}=\frac01$$ we get $$\frac{s_1\cdots s_nb^n+sa_1s_2\cdots s_nb^{n-1}+\cdots+s^na_ns_1...s_{n-1}}{s^ns_1\cdots s_n}=\frac 01,$$ so there is $u\in S$ such that $$u\left(s_1\cdots s_nb^n+sa_1s_2\cdots s_nb^{n-1}+\cdots+s^na_ns_1...s_{n-1}\right)=0.$$ (It seems the book missed this part.) Now multiply the equation by $(us_1\cdots s_n)^{n-1}$ and find that $us_1\cdots s_nb$ is integral over $A$.