Let S be the class of all rings R which have 1 and satisfy this condition:


for every "non-zero" right ideal I of R there exists a "proper" right ideal J of R such that I + J = R. (The + here is not necessarily direct.)


All semisimple rings are in S and (commutative) local rings which are not fields are not in S. The ring of integers Z is also in S and so S properly contains the class of semisimple rings.


My questions:


Will this condition by itself force an element of S to have any (known, interesting) structure?


A more important question: 


What about simple rings which are in S? For example, do they have to be semisimple? (Unlikely!) 



Semisimple-ish rings!
By Zorn's lemma, each right ideal is contained in a maximal right ideal,
therefore if $I+J = R$ then $I+M = R$ where $M$ is a maximal right ideal.
If $I+M\ne R$ for all maximal right ideals $M$ then $I\subseteq M$ for
all maximal ideals $M$. Thus $I\subseteq J(R)$, the Jacobson radical of $R$
which is the intersection of all maximal right ideals of $R$. Hence
condition $S$ is equivalent to $J(R)=0$.


A ring with vanishing Jacobson ideal is called *semiprimitive*.
As $J(R)$ is also the intersection of the maximal left ideals of $R$
then the property of semiprimitivity is left-right symmetric.
There are plenty of examples of semiprimitive rings which are not
semisimple. For instance every simple ring is semiprimitive
and every subdirect product of semiprimitive rings is semiprimitive
($\mathbb{Z}$ is a subdirect product of finite fields).
As a reference see Section 10.4 of P. M. Cohn
*Algebra* (2nd ed. vol 3) Wiley 1991.


