Let $A$ be a domain and $J\subset A$ be an ideal. Let $m$ be a maximal ideal in $A$, and $I=mJ$. Let $x\in J\setminus I$. I want to show that $J=xA+I$.
As pointed out by Mindlack's answer, above statement is wrong. The full setup in the book is as the following:
Let $A$ be a Dedekind domain, then an ideal $I$ admits a unique factorizaiton by primes $I=p_1^{e_1}\dots p_r^{e_r}$. Let $J=p_1^{e_1-1}\dots p_r^{e_r}$. Let $x\in J\setminus I$. I want to show that $J=xA+I$.
The statement is used to prove that $A/I\cong (S^{-1}A)/I (S^{-1}A)$ for a multiplicative set $S$ disjoints from each $p_i$ in decomposition of $I$. So could you please not use that conclusion, to avoid circular argument?
Edit: I'm answering the modified question, my point about the original question is below. Note that by the Chinese remainder theorem $A/I \cong \prod_i{A/p_i^{e_i}}$, so $J/I \cong p_1^{e_1-1}A/p_1^{e_1} \cong p_1^{e_1-1}A_{p_1}/p_1^{e_1}A_{p_1}$ but $A_{p_1}$ is a DVR (thus a PID) and thus the above is a one-dimensional $A/p_1$-vector space, so is generated (over $A$) by the image of any nonzero element, for instance the image of $x$.
The original question is wrong in general. Indeed, assume the result holds with $J=m$ and $A$ Noetherian. Then $m/m^2$ as a vector space over $A/m$ is generated by $[x]$ so has dimension one. As $A$ is an integral domain, $A_m$ (thus $A$) is a field or $A_m$ is a regular local ring of dimension $1$ hence a DVR.
If you want an explicit example, take $A=\mathbb{C}[X,Y]$, $m=J=(X,Y)$. Then $Y \in m \backslash (XA+m^2)$.