My approach: Suppose I know that $aR = R$; $a \in R$, $\implies Ra = R$ for a semisimple ring $R$
Now I think I can apply this result as follows. Since $I$ is an ideal in $R$, it can be considered as a ring in itself ($R = I \oplus U , \phi: R \rightarrow I$ a ring hom with kernel $U$) and we know that $a \in I$, therefore in the ring $I$, $aI = I$. Now using the above result that $aR = R \implies Ra =R$ , I conclude that $Ia = I$ and hence $I \subseteq Ra \subseteq R$. But $RaR = I$; since $aR = I$ and $Ra \subseteq RaR$ therefore $Ra = I$
But for some reason my proof doesn't seem convincing enough, so can somebody crosscheck it once again and provide solution (hint) if its wrong.
It is a valid idea, but I think the last part is not very clear.
If $aR=I$ is an ideal, then it is immediate that $Ra\subseteq I$. It only remains to be shown that $I\subseteq Ra$.
From the lemma, $aI=I$ implies $I=Ia\subseteq Ra$ and you are done.
The lemma you suggested is easy to prove after all since $aR=R$ implies $a$ is right invertible, hence left invertible in a semisimple ring. This it is a unit and $Ra=R$ too.