Let $R$ be an integral domain. Suppose that for any two nonzero ideals $I$ and $J$, we have $I \oplus J$ is isomorphic to $R \oplus IJ$ as $R$modules. Does this implies $R$ is a Dedekind domain?

As Sampath has pointed out, we may assume that $R$ is local. Your hypothesis implies for any two nonzero ideals $I,J$, you have a surjection onto $R$. By Nakayama, this implies either $I\to R$ or $J\to R$ is surjective, since if neither is, then the both have images contained in the maximal ideal and so does their sum. But this means one of them is principal. Having recognized the confusion I caused by my terseness, let me be more explicit. First, my definition of DD is: R a domain (not a field) and for any nonzero ideal $I$, there exists another ideal $J$ such that $IJ$ is isomorphic to $R$. Easy to see that $\text{Hom} (I,R)$ for any nonzero ideal $I$ can be identified naturally with $J=\{x\in KxI\subset R\}$ where $K$ is the fraction field of $R$. Then, the definition of DD means that for any nonzero ideal $I$, defining $J$ as above, $IJ=R$. Now assume that for any nonzero ideal $I$ of $R$, there exists a surjection $I\oplus I$ to $R$. Then, I claim that $R$ is a DD. The hypothesis implies, there exists $a,b\in I$, $x,y\in K$ with $xI,yI\subset R$ and $xa+yb=1$. Easy to check then that $I$ is generated by $a,b$. Thus all ideals are generated by atmost two elements and in particular $R$ is Noetherian. Now, by the above localization argument, $I$ is locally principal and the rest is clear. I hope this is clearer. Even without localizing, letting $J$ as above, we have $x,y\in J$ and thus $1\in IJ\subset R$ and hence $IJ=R$. 

