Suppose we have a ring $R$ and a multiplicative set $S$. Can we define $S$ rings analogous to Group Rings?
I think we'll have some problems if inverses do not exist in $S$, since when multiplying two linear combinations, we get this step in between
$$ \sum_{y \in S} \sum_{x \in S} a_{x} b_{y} (xy) $$
When $S$ is a group, this is equal to
$$ \displaystyle \sum_{t \in S} \left(\sum_{x \in S} a_{x} b_{x^{-1}t}\right) t = \sum_{t \in S} \left(\sum_{xy=t} a_{x} b_{y}\right) t $$
and for that we'll need existence of inverses.
But I'm not sure since many books leave out the intermediate step and directly write the product as the latter formula.
Can someone tell if we can define an analogue to Group Rings for Multiplicative Sets?
For any semigroup $S$, you can form the set of formal linear combinations and define multiplications exactly the same way as for group rings:
$(\sum r_x x)(\sum r_y y) = \sum_{z\in S\,,\, z=xy} r_xr_y z$
That is, you just multiply everything out using distributivity, then group the terms by the resulting $xy$ product results.
The thing you quoted that uses inverses is equivalent when the set is a group, but it is not necessary at all to define the product.
This is called the semigroup ring and is pretty well-studied, although not so well-studied as group rings. The resulting ring need not have a multiplicative identity element.
If you had in mind a nonempty multiplicatively closed subset of a ring, that would indeed form a semigroup. You just need the multiplication to be associative to give the formal linear combinations a ring structure.