$R$ is a commutative ring. $S^{-1}R$ is a ring of quotients and $S^{-1}I$ is an extension of ideal $I$ in $S^{-1}R$. In Algebra by Hungerford, Theorem III.4.8, page 146, he says when $R$ is a commutative ring with identity, $S^{-1}I = S^{-1}R$ if and only if $S \cap I \neq \emptyset$. The proof is:
If $S^{-1}R = S^{-1}I$, then $\varphi _{S} : R \rightarrow S^{-1}R, r \mapsto rs/s$ and $\varphi_S^{-1} (S^{-1}I) = R$ whence $\varphi_S(1_R) = a/s$ for some $a \in I, s \in S$. Since $\varphi_S (1_R) = 1_Rs/s$ we have $s^2s_1 = ass_1$ for some $s_1 \in S$. But $s^2s_1 \in S$ and $ass_1 \in I$ imply $S\cap I \neq \emptyset$.
If I go this way:
Since $S^{-1}R = S^{-1}I$ there is $ a \in I$ and $a/s = s/s$ so $ss_1a = s^2s_1 \in I$ and $S$. Therefore $I \cap S \neq \emptyset$.
Then we can draw the conclusion without $R$ to have an identity.
Am I right or something wrong?
If $S^{-1}I=S^{-1}R$, then you can choose $r\in S$, and by assumption $$ \frac{r}{r}=\frac{x}{s} $$ for some $x\in I$ and $s\in S$. This means that there is $t\in S$ such that $rst=xrt$. This element belongs to $S\cap I$.
Conversely, if $x\in S\cap I$, then $x/x\in S^{-1}I$. This element is the identity in $S^{-1}R$, so $S^{-1}I=S^{-1}R$.
No identity in $R$ is assumed.
Your proof is correct as well.