Can $Rx$ be contained in $I$ for some nonzero $x\in S$, if $I:=\langle R\setminus S\rangle$?

Question

Let $R$ be any ring and let any set of elements $S\subset R$ (proper) be such that $0_R,1_R\in S$ with some other elements of $R$. Now, define $$I:=\langle R\setminus S\rangle=\langle\{a\in R: a\notin S\}\rangle.$$ That is, $I$ is generated by the set of elements not in $S$.

Question: Is there a possibility some left ideal $Rx$ to be contained in $I$ for some nonzero $x\in S$?

Answer 1

Let $R=\mathbb{Z}$ and $S=\mathbb{Z}\setminus\{2\}$. Then $I=2\mathbb{Z}$ and $4\in I\cap S$.

Answer 2

Let $R=\mathbb{R}^{3\times 3}$ and let $S=\left\{E_{i,j}\mid 1\leq i,j\leq 3\right\}\cup \left\{0_R,1_R\right\}$ where $E_{i,j}$ is the matrix with zeroes everywhere and a $1$ at position $(i,j)$. Then $I=\left\langle R\setminus S\right\rangle=R$. Clearly $RE_{1,1}$ is contained in $I$.