Can two different multiplicative systems give same localisation? Can we have two different multiplicative systems $S_1$ and $S_2$ in $\mathbb{Z}$ having same localisations $\mathbb{Z}_{S_1}=\mathbb{Z}_{S_2}$? 
I have a trivial solution by taking two different subrings and adjoining $1$ to them gives required $S_1,S_2$ but that collapses the localizations to trivial rings. I wanted something more interesting. Considering multiplicative systems without $0$...
 A: Here is a very general criterion.  Let $A$ be a commutative ring and let $S\subseteq A$ be a multiplicatively closed subset.  Define the saturated closure of $S$ to be $$\operatorname{sat}(S)=\{a\in A: a\text{ divides some element of }S\}.$$ Note that $\operatorname{sat}(\operatorname{sat}(S))=\operatorname{sat}(S)$ so this really is a "closure" operation.

Theorem: Let $S,T\subseteq A$ be multiplicatively closed subsets of $A$.  Then the localizations $A_S$ and $A_T$ are isomorphic as $A$-algebras iff $\operatorname{sat}(S)=\operatorname{sat}(T)$.

This gives lots of examples.  For instance, in the case $A=\mathbb{Z}$, you can take any nonzero integer $n$ and let $S\subset\mathbb{Z}$ be the multiplicatively closed subset subset generated by $n$.  Then $\operatorname{sat}(S)$ will be the multiplicatively closed subset generated by the prime factors of $n$.  So if $m$ is any other integer with the same set of prime factors as $n$ and $T$ is the multiplicatively closed subset generated by $m$, then $\mathbb{Z}_S\cong\mathbb{Z}_T$.
Proof of Theorem: Note that in any commutative ring, any divisor of a unit is a unit.  It follows that for any $A$-algebra $B$, every element of $S$ is a unit in $B$ iff every element of $\operatorname{sat}(S)$ is a unit in $B$.  Thus the localizations $A_S$ and $A_{\operatorname{sat}(S)}$ have the same universal property as $A$-algebras, and so they are isomorphic.  It follows immediately that if $\operatorname{sat}(S)=\operatorname{sat}(T)$ then $A_S\cong A_T$.
Conversely, I claim that $\operatorname{sat}(S)$ is exactly the set of elements of $A$ which are units in $A_S$.  By the discussion above, every element of $\operatorname{sat}(S)$ is a unit in $A_S$.  Conversely, suppose $a\in A$ is a unit in $A_S$.  This means there exists $b\in B$ and $s\in S$ such that $\frac{a}{1}\cdot\frac{b}{s}=\frac{1}{1}$ in $A_S$, which means there exists $t\in S$ such that $t(ab-s)=0$.  But then $abt=st$ and so $a$ divides the element $st\in S$, so $a\in\operatorname{sat}(S)$.
It follows that if $A_S\cong A_T$ then $\operatorname{sat}(S)=\operatorname{sat}(T)$, since the elements of $A$ that are units in $A_S$ and $A_T$ are the same.
