That's an OK definition of "prime" but I find another one (product of two right ideals is zero iff one of them is already zero) useful to keep in mind.
A prime ring is always indecomposable: if $R=S\oplus T$, then $S$ and $T$ are ideals of $R$ whose product is zero.
I can't offer any support for a sort of converse, because there does not seem to be a strong connection. There is not even a good result for commutative Artinian rings. A commutative Artinian ring is indecomposable iff local, but a commutative Artinian ring is prime iff it is a field.
Any local Artinian ring which is not a division ring is indecomposable (thanks to the absence of any idempotents except 0 and 1) but it isn't prime because its radical is a nonzero nilpotent ideal. This includes your example, which is Artinian and local.
Let's say you wanted to show that nonprime rings with condition X are decomposable. You could reason: "There exist nonzero ideals $A$ and $B$ such that $AB=0$. Because of X there exists a nontrivial central idempotent of $R$." Maybe someone can think of a clever condition X?