Comments on the partial progress.
Unfortunately, this is a case of something that sometimes happens with students: a strong condition was given (semisimple Artinian) and then you decided to quickly drop down to something weaker (Artinian/Noetherian) possibly because you feel more comfortable with that condition.
It is possible to prove that if $G$ is infinite then $R[G]$ can't be Artinian, but as I alluded to in the comments it's not easy enough to be a homework problem.
If you start with assuming $G$ is infinite and then try to show that $R[G]$ isn't Noetherian, you are unfortunately doomed to failure, because (I think I am recalling correctly) there are examples of infinite groups $G$ such that $R[G]$ is Noetherian.
This highlights some of the dangers of "simplifying" givens too quickly. We can prove it in the following way taking full advantage of supposed semisimplicity of $R[G]$. Specifically, we'll use the fact that ideals are summands.
- Let $A$ be the augmentation ideal of $R[G]$. That is, it is the kernel of the projection $R[G]\to R$ which sends $g\mapsto 1$ for all $g\in G$. Show that $1-g\in A$ for all $g\in G$.
- Suppose $R[G]$ is semisimple. Then $A$ is a summand of $R[G]$, and in particular we can find an idempotent $e\in R$ such that $R[G]e=A$ and $R[G]e\oplus R[G]f=R[G]$ where $f=1-e$ is another idempotent. Prove that $(1-g)f=0$ for every $g\in G$ using the previous point. Consequently, $f=gf$ for all $g\in G$.
- $f$ has to be nonzero, since the augmentation ideal is proper. Let $h\in G$ be a term of $f$ that has a nonzero coefficient. Show using the previous point that $gh$ appears with a nonzero coefficient in the expression of $f$.
- Since every element of $G$ can be expressed as $gh$ for some $g$, this means that all elements of $G$ have nonzero coefficients in the expression of $f$... but it isn't possible for $f$ to have infinitely many nonzero terms. Thus $G$ isn't infinite.