Does there exist a ring such that all modules are decomposable?
If we consider an infinite dimensional $k$-linear space $V$, then $V\cong V \oplus V $, then we can show that $R \cong R \oplus R$ as $R$-modules, where $R: = End_k(V)$ (this is an example from Rotman <An introduction to homological algebra> Example 2.36). We know $R$ is a von Neumann regular ring which does not have IBN. Maybe $R$ satisfy the condition but I can't prove it.
Not for rings with identity, because all rings with identity have simple modules which are indecomposable. (These are found by applying Zorn's Lemma to the proper right ideals of $R$ and finding a maximal right ideal $T$, after which $R/T$ is a simple right $R$ module.)
This is certainly the case for the ring $R$ which you have chosen, since endomorphism rings always have the identity endomorphism.