Your "trivial" examples all resulted from direct sum decompositions of the ring $R$.  By asking for examples without idempotents, you are asking for rings that do not have direct sum decompositions.  In a noncommutative ring $R$, the corresponding would be a ring that has no *central* idempotents.  I can provide a noncommutative example that is "in-between," so that it has nontrivial idempotents, but no nontrivial central idempotents.

For an ideal $J$ in a noncommutative (read: not-necessarily-commutative) ring $R$, there is a way to reformulate when the right $R$-module $R/J$ is flat.  In T.Y. Lam's *Lectures on Modules and Rings*, Proposition 4.14 implies that $R/J$ is right flat if and only if, for every left ideal $_RL \subseteq R$0, 
$$
J \cap L = JL.
$$
(Notice that this provides an alternative way to verify that for such $J$, $J^2 = J$.)

Now given a field $k$ (or even a division ring!), let $V$ be a (right) vector space of countably infinite dimension, and let $R = \operatorname{End}_k(V)$, acting on $V$ from the left.  This ring has many idempotents, corresponding to direct sum decompositions of $V$.  One can show that $R$ has precisely three ideals, namely $0$, $R$, and the ideal $J$ consisting of endomorphisms of finite rank (see Exercises 3.15-3.16 of Lam's *Exercises in Classical Ring Theory*).  In particular, $R$ does not decompose as the direct sum of two nontrivial subrings.  Let $f$ be any finite-rank element of $R$, and let $p$ in $R$ be a projection of $V$ onto the image of $f$.  Certainly $f = pf \in Jf$.  This makes it easy to show that $J$ satisfies $J \cap L = JL$ for every left ideal $L$ of $R$, and it follows that $R/J$ is flat.