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.
