There is an alternative argument for the free group; not using that free groups are residually finite-dimensional.
Let $\pi$ be a faithful representation of $C^{\ast}(F)$ on a Hilbert space $H$. Then, as $U(H)$ is connected, $\pi$ can be deformed to the trivial representation in the point-norm topology, i.e. there exists a family of unitary representations $\pi_t$ for $t \in [0,1]$, such that $t \mapsto \pi_t(a)$ is norm-continuous for each $a \in C^{\ast}(F)$, $\pi_0=\pi$ and $\pi_1(g)=1_H$ for all$ g \in F$.
Now, if $ab=1$ in $C^{\ast}(F)$, then $\pi_t(ba)$ is a continuous path of projections ending at $1_H$. Hence, $\pi_0(ba)=1_H$ and $ba=1$, as $\pi$ was faithful.
EDIT: The same argument works if the $C^{\ast}$-algebra embeds into some contractible algebra (i.e. homotopy equivalent to $\mathbb C$). However, even though many reasonable toplogical spaces are quotients of contractible topological spaces, only few reasonable $C^{\ast}$-algebras have this property. There is a close relationship with the concept of quasi-diagonality, which appeared in the work of Voiculescu.
