If you require the distributive law on just one side, you obtain what is called a near ring.
If you require distributivity on both sides, this tends to force $+$ to be commutative.
Indeed, computing the product $(\alpha + \beta)(a + b)$ using distributivity on the left, and then on the right, and then doing it again in the opposite order, you deduce that
$$\beta a + \alpha b = \alpha b + \beta a.$$
So at least on the additive subgroup of $R$ generated by elements which are products of
two other elements, the operation $+$ is commutative. If e.g. you require in addition that $R$ contains a multiplicative identity, then $+$ will be commutative on all of $R$.
Edit: This argument is a variant of the Eckmann-Hilton argument.
It might also help to think about the basic example of a near-ring (as discussed in wikipedia), namely maps from a group $G$ to itself, with $+$ being the group operation (applied pointwise to maps) and $\cdot$ being composition of maps. The right distributive law is trivially true, but imposing the left distributive law would then say that we are looking at maps of a group that preserve the group operation, i.e. endomorphisms from $G$ to $G$. But for
this to be a near-subring of the near-ring of all maps, then we would need that the pointwise sum of two homomorphisms is again a homomorphism, a condition which holds only if $G$ is abelian.