$ba\in Ra$, therefore $Rba\subseteq Ra$ (it has to absorb on the left.)
There is no reason for $Rba\subseteq Rb$. You would have to show, in particular, that $ba\in Rb$. Without commutativity, this is not a sure thing.
For a counterexample, you can look at $R=\mathbb Z\langle x,y\rangle/(y^2, xy)$, and examine the left ideal $Ry$ and whether or not $xy$ is in it.