After answering, the question changed so I adapt my answer to the new question.

As I was remarking in the comments to your question, it is impossible to construct such ring. In fact, as you want that, for all $x,y\in R\setminus\{0\}$, the intersection $\{r\in R:xr\in (I:I)\}\cap\{r\in R:xry\notin I\}\neq \emptyset$, in particular you need that
$\{r\in R:xry\notin I\}\neq \emptyset$ for all $x,y\in R\setminus\{0\}$. Notice that, if $x\in I$, then, as $I$ is a right ideal, $xry=x(ry)\in I$ for all $r$ and $y\in I$, so this set is always empty.

Furthermore, if you have $r$ such that $xr\in (I:I)$, then, by definition of $(I:I)$, $xry\in I$ provided $y\in I$. So, if $y\in I$, then $\{r\in R:xr\in (I:I)\}\cap\{r\in R:xry\notin I\}= \emptyset$.

Maybe you have some hope looking in a non-commutative ring and taking $I$ to be a right **but not left** ideal and imposing your second condition just for $x,y\in R\setminus I$. Furthermore, another necessary condition is that $I\neq (I:I)$. Anyway, this is another question!

Finally, let me also add that the first condition $(I:I)\cap \{t\in R:t(I:I)t\subseteq I\}\neq \{0\}$ is implied by $I\neq\{0\}$ in fact $I$ is always contained in this intersection.

for all$x,y\in R$? In particular, if $x\in I$, then, as $I$ is a right ideal, $xry$ belongs to $I$ for all $r$ and all $y$. So... for $x\in I$, the set {$r\in R|xry\notin I$} is always empty. For example, take $x=0$ or $y=0$, you will see that your second condition is impossible to satisfy. So the answer is "NO!" there is no such ring:) – Simone Virili Nov 27 '12 at 11:03