What's to stop you from just freely building a counterexample? That is, let the ring be generated by elements $a,b_1,b_2,\dots,b_n,\dots$ subject to (only) the relations $a=x^nb_n$ for all positive integers $n$. Then $a$ is in $I(x)$. Unless I'm making a stupid mistake, the only solutions $q$ of $a=xq$ are finite linear combinations (with integer coefficients adding to 1) of the elements $x^{n-1}b_n$, and none of those are in $I(x)$.

What's to stop you from just freely building a counterexample? That is, let the ring be generated by elements $x,a,b_1,b_2,\dots,b_n,\dots$ subject to (only) the relations $a=x^nb_n$ for all positive integers $n$. Then $a$ is in $I(x)$. Unless I'm making a stupid mistake, the only solutions $q$ of $a=xq$ are finite linear combinations (with integer coefficients adding to 1) of the elements $x^{n-1}b_n$, and none of those are in $I(x)$.