The difference between $l^1(G)$ and the reduced group $C^*$ algebra $C_r^*(G)$

In Q2 a couple of things are mixed up. If you consider $ell^1 G$ with the point-wise multiplication, then $ell^1 G$ is a commutative non-unital Banach algebra and the closure of the space of commutators (with respect to the multiplication which is induced by the group multiplication) is indeed a two-sided ideal. The quotient is given by $ell^1$-functions on the space of conjugacy classes.

However, if you consider $ell^1 G$ only with the multiplication coming from the group, then it is a unital (typically) non-commutative Banach algebra and the ideal generated by the commutators is larger that just the closure of the space of commutators. The quotient is $ell^1 H$, where $H$ is the abelianization of $G$.

For the reduced $C_{red}^*$-algebra, the ideal structure can be quite different compared with $ell^1 G$. For example, if $G$ is a non-abelian free group, then $C_{red}^star G$ is simple and there is only the trivial quotient. In particular, the ideal generated by the commutators is everything. However, if $G$ is amenable, the quotient by the commutator ideal can be identified with the reduced $C_{red}^star$-algebra of the abelianization of $G$.

Another way to read your question is the following: Can an element of finite order in $G$ become equal to an element of infinite order modulo commutators in $C_{red}^star G$? That is now a question about traces on group $C_{red}^*$-algebras. Equivalently, you could ask: Is there a trace on $C_{red}^star G$ which distinguishes element of finite order from those of infinite order. For many group (for example a free product of finite groups; not both $C_2$) there exists a unique trace on $C_{red}^star G$. In particular, traces cannot distinguish between non-trivial conjugacy classes.

If $G$ is amenable, the situation is different. I do not know whether the reduced $C^star$-algebra of an amenable group supports sufficiently many traces in order to distinguish conjugacy classes.