If $I$ is a maximal ideal and $a\in R -I$, then the assumption that $I + (a) = R$ gives a contradiction While I'm trying to prove that

Let $S$ be a multiplicative set in the commutative ring $R$ with
  identity s.t $0 \not \in S$.Let $I$ be a maximal ideal in $S^c = R -
 S$. Then show that $I$ is a prime ideal.

.I stuck at showing that 

For $ab \in I$ and $a,b \not \in I$, if $a \in S^c - I$, then $$I +
 (a) = R$$ gives a contradiction.

My first (and in fact only) attempt was observing that $\exists (j \in I, r \in R)$ s.t 
$$j + ra = 1_R.$$
But I couldn't find anything wrong by with such a claim. Of course, this might have throw any contradiction, but I'm not sure, so my question is that do we have any contradiction in here, if so, how to show it ?
 A: A contradiction? No.
Observe that $I+(a)$ is an ideal and is larger than maximal ideal $I$ because it contains $I$ as a subset and contains $a\notin I$ as an element. 
Then $I+(a)=R$ is the only possibility, because $I+(a)\neq R$ would violate the maximality of $I$.
It is not a contradiction but a correct conclusion.

P.S.
This answer was posted before  the illuminating edit of the OP.
A: With the added context, we can now see where your problem lies:

While I'm trying to prove that
Let $S$ be a multiplicative set in the commutative ring R with identity s.t $0∉S$. Let I be a maximal ideal in $S^c=R−S$. Then show that $I$ is a prime ideal.

The goal is to show that if $I$ is an ideal maximal with respect to containment in $S^c$, then $I$ is prime.  In fact, $I$ is not necessarily a maximal ideal, it's just maximal with respect to the property I just stated.
That seems to be your misunderstanding. You need to proceed instead with $(a,I)\cap S\neq\emptyset$, rather than $(a,I)=R$. Also consider further assuming $(b,I)\cap S\neq\emptyset$ and see what you can do with the two statements.
A: This is not a contradiction. Show that $(2)$ is maximal (the quotient is a field!) in $\mathbb Z$ and that $(2)+(3)=\mathbb Z$ (show that $1$ is in the sum.)
In fact, if $I \subset A$ is a maximal ideal and $I+(a) \neq A$ for $a \notin I$, we have that $I \subset I+(a) \subsetneq A$ so how could $I$ have been maximal?
A: As mentioned above, this is not a contradiction. 
If you're trying to prove that $I$ is NOT a maximal ideal, I think what you need to say is that there exists some $a \in R-I$ such that $I+(a) \subsetneq R$, which probably can be achieved by yielding a contradiction from $j + ra = 1_R$.
