What you did for the left inverse is fine, but what you did for the right inverse doesn’t work: you know that $AR=I$, but you don’t know that $RA=I$, so you can’t say that $x=Rb$. (Actually, you may have seen that $ARb=Ib=b$ and realized on that basis that $Rb$ is a solution; if so, you’re right, but that doesn’t say anything one way or the other about whether it’s the only solution. In any case, you should justify it.)
Suppose that $A$ is an $m\times n$ matrix with a right inverse $R$. Then $R$ is $n\times m$, and $AR$ is $m\times m$. This implies that the $m$ rows of $A$ are linearly independent (why?), i.e., that the rank of $A$ is $m$, and hence that $m\le n$. From here there are several ways to argue that $Ax=b$ has at least one solution, depending on what you’ve already proved. One very straightforward way, however, is to consider what happens to the augmented matrix $\begin{bmatrix}A&b\end{bmatrix}$ when you row-reduce it.