Replacing the sum of two elements with a single element in a non-commutative ring I would like to know if the following is true in a non-commutative ring $R$,
For every $r_1,r_2,s_1,s_2,a \in R$, there exists $r^*,s^* \in R$ such that $r^*as^* = r_1as_1 + r_2as_2$.
I certainly know it is possible for one-sided cases, i.e, there exists $r^*,s^* \in R$ such that $r^*a = r_1a + r_2a$ and $as^* = as_1 + as_2$ by choosing $r^* = r_1 + r_2$ and $s^* = s_1 + s_2$ due to the distributive property.
I first tried using $(r_1+r_2)a(s_1+s_2) = r_1as_1 + r_1as_2 + r_2as_1 + r_2as_2$, but do not think $r_1as_2 + r_2as_1 = 0$, so I am stuck.
Any help would be greatly appreciated, thanks.   

Your claim is false. Let $R$ be the ring of linear maps from the Euclidean plane in itself. Let $a$ be the projection on the $x$ axis, and let $b$ be the simmetry wrt $y=x$. Then $a + bab$ is the identity, while $ras$ has rank at most 1 for any choice of $r, s$.
