For some choices of $R$ there are no other multiplications on $R$, even if you require that they satisfy no axioms besides distributing over addition. For instance, consider $R=\mathbb{Q}/\mathbb{Z}$ and suppose you have a product on $R$ that distributes over addition. Let $x,y\in\mathbb{Q}/\mathbb{Z}$ be arbitrary; say $x$ is represented by a rational number $\frac{a}{b}$ and $y$ is represented by a rational number $\frac{c}{d}$ (with $a,b,c,d\in\mathbb{Z}$). Let $x'$ be the element of $\mathbb{Q}/\mathbb{Z}$ represented by $\frac{a}{bd}$. Then $dx'=x$ (where $dx'$ just means a sum of $d$ copies of $x'$), so since our product distributes over addition we have $$xy=(dx')y=d(x'y)=x'(dy).$$ But $dy$ is represented by the rational number $d\frac{b}{d}=b$ which is an integer, so $dy=0$. Thus $xy=x'\cdot 0=0$.

(More generally, a similar argument applies to any abelian group $R$ which is divisible but in which every element has finite order.)