What happens if we remove the requirement that $\langle R, + \rangle$ is abelian from the definition of a ring? Ever since I learned the definition of a ring, I've wondered why the additive group is required to be abelian.  What happens if we allow $\langle R, + \rangle$ to be nonabelian as well as $\langle R, \cdot \rangle$?  Is this impossible?  If not, what is the this type of algebraic structure called?
Here is the definition of a ring that I am using:

  
*
  
*$\langle R , + \rangle$ is an abelian group with identity $0$.
  
*$\langle R , \cdot \rangle$ is associative.
  
*$a\cdot (b + c) = a\cdot b + a \cdot c$.
  
*$(a + b)\cdot c = a\cdot c + b \cdot c$.
  

 A: There is an algebraic structure called a near ring that allows for the "addition" to be non-commutative, though it only requires distributivity on the right.
Thus, if we take the ring axioms as you've listed them, and change the first one to only require that $\langle R,+\rangle$ is a group, we get the axioms for a right near ring that is also a left near ring under the same operations.
A: 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.
