Let $A$ be a commutative ring with unity $1$ different from $0$ and $S\subset A$ be a multiplicatively closed. (Edited after Answer/comment by Bernard: $S$ is multiplicatively closed means whenever $x,y\in S$ then $xy\in S$; and we take $S\neq \phi$.)
On set $\{\frac{a}{s}:a\in A, s\in S\}$ we define equivalence relation $\frac{a}{s}\sim \frac{a'}{s'}$ iff $(as'-a's)t=0$ for some $t\in S.$ On the set $A_S$ of equivalence classes, we define usual additions and multiplications as in $\mathbb{Q}$ we do.
My questions are simply whether following statements are correct?
1. $A$ will be ring with unity different from $0$ only if $1\in S$.
2. Suppose $1\notin S$. Let $S^*=\{1\}\cup S$. Then $A_S$ and $A_{S^*}$ are isomorphic rings (i.e. there is bijection $f:A_S\rightarrow A_{S^*}$ which is additive homoorphism and preserves multiplication).
My exact doubts:
(a) I didn't find any reason to include/exclude $1$ in defining localization, which forced me for true/false testing statement 1
(b) In the case when $A$ is integral domain, the answer of Q.2 is Yes (it is an exercise in Janusz's Algebraic number theory.) But my question is - if $A$ not integral domain but just commutative ring with unity different from $0$, is statement 2 correct?
For question 2, $A_S$ and $A_{S^*}$ are solutions of the same universal problem (making the elements of $S$ and the elements of $S\cup\{1\}$ invertible are equivalent problems), so they're isomorphic.
On the other hand, it depends on what is your definition of a multiplicatively closed set. With Bourbaki's definition:
your question is meaningless.