Suppose that for every nonzero $R$-ideal $I$ and element $a \in I$ there exists a unique $R$-ideal $J$ such that $IJ=(a)$. Then $R$ is Noetherian. 
Suppose that for every nonzero $R$-ideal $I$ and element $a \in I$ there exists a unique $R$-ideal $J$ such that $IJ=(a)$. Then $R$ is Noetherian.

I'm having trouble proving this. To note, $R$ is assumed to be a commutative ring here.  
Consider $I_1 = (a_1)+(b_1)$ and $I_2 = (a_2)+(b_2)$.  Is it true that $I_1I_2 = (a_1a_2)+(a_1b_2)+(b_1a_2)+(b_1b_2)$?
If so, then the property specified in the question means that if $I_1$ contains multiple generators, of which $a_1$ is one, there exists $I_2$ with generators $(a_2)+(b_2)+\dots$ such that $(a_1) = I_1I_2 = (a_1a_2)+(a_1b_2)+(b_1a_2)+(b_1b_2)+\dots$, right?  
I'm not really sure where I'm going with this.  My original intention was to assume that $R$ was not Noetherian, i.e. it has an infinitely generated ideal, and then show a contradiction.  
Something else I was trying was to show that for any ideal $I_0$ with generators $(a_1)+(a_2)+\dots$ one can write $I_0 = (a_1)+(a_2)+\dots = I_0I_1 + I_0I_2 + \dots = I_0(I_1+I_2+\dots)$ where $I_i$ is the unique ideal such that $I_0I_i = (a_i)$.  I'm not sure where to take this either.  
Any hints or explanations would be greatly appreciated.  
 A: I'm not sure how helpful this will be, but this is too long for a comment. 
The problem puts me in mind of Dedekind domains, for which there are several definitions. Notice that $R$ is in fact an integral domain (otherwise if $bc = 0$ and we take $I = (b)$ and $a = 0 \in I$, we would have $(b)(0) = (0)$ and $(b)(c) = 0$ which destroys uniqueness of $J$). 
Under the EOM account, an integral domain $R$ with field of fractions $K$ is a Dedekind domain iff the monoid of fractional ideals is a group. A fractional ideal is an $R$-submodule of $K$ of the form $r^{-1}I$ where $r \in R$ is nonzero and $I$ is an ideal of $R$, and multiplication of fractional ideals is defined similarly to multiplication of ordinary ideals; the unit element is $R$. Ordinary ideals form a submonoid of the monoid of fractional ideals, and it looks as though the hypothesis of the problem has to do with cancellability in this submonoid, so that fractional ideals are obtained by adjoining formal inverses to this cancellable monoid and thus form a group. Under this line of thinking, the hypothesis would imply that $R$ is a Dedekind domain. On the other hand, Dedekind domains under another definition are a fortiori Noetherian. 
I realize this is not directly helpful, but you might want to consult Zariski-Samuel's text on commutative algebra to see what they say on Dedekind domains. (Unfortunately I can't find my copy at the moment to verify this is a good place to go). 
A: As @user43208 pointed out, the hypothesis implies that $R$ is an integral domain. We denote by $K$ its field of fractions.
A fractional ideal (in $\boldsymbol{K)}$ is by definition a $R$-submodule of $K$, say $I$, such that $bI\subseteq R$ for some $b\in R\setminus0$. We say that such fractional ideal $I$ is invertible if there is another fractional ideal $J$ such that $IJ=R$, where $IJ:=\{\sum i_kj_k: i_k\in I, j_k\in J\}\,.$ There is only one possibility for the inverse $J$ of the fractional ideal $I$, namely $I^{-1}:=\{z\in K: zI\subseteq R\}$: in fact, if $J$ is a fractional ideal satisfying $IJ=R$, then both $J\subseteq I^{-1}$ and $I^{-1}I\subseteq R$ hold by definition of $I^{-1}$, so  $J(I^{-1}I)\subseteq JR$, that is $I^{-1}\subseteq J$. It is clear that a (finite product) of fractional ideals is invertible iff each factor is invertible. Finally, every nonzero cyclic $R$-submodule of $K$ is an invertible fractional ideal: if $I=Ra$, with $a\in K^\ast$, then $I(Ra^{-1})=R$.
Let $I$ be a nonzero ideal in $R$, and let $a\in I\setminus0$ be fixed. Let $J$ be the unique ideal in $R$ such that $IJ=Ra$. Obviously the ideals in $R$ are fractional ideals in $K$, and since $Ra$ is invertible, then $I$ is invertible too, so we have $I^{-1}I=R$. Therefore there are finitely many $i_k\in I$ and $a_k\in I^{-1}$ such that $\sum a_ki_k=1$, so for any $r\in I$ we have $r=\sum (ra_k)i_k$. Since $ra_k\in R$ for each $k$ (because $r\in I$ and $a_k\in I^{-1}$), this proves that $I=\sum Ri_k$ is finitely generated, so $R$ is Noetherian as desired.
