Groups for which all projections of $C^*_{\text{red}}G$ belong to $\mathbb{C}G$ Revision:  According to comment of Wojowu we give a complete revise for this post.
A group $G$ is a pr-group if all projections of $C^*_{\text{red}} G$ are contained in its dense subalgebra $\mathbb{C}G$.
What is a counterexample of this situation?
Is there a characterisation of pr-groups?
 A: If the group is torsion-free, pr-ness is clearly implied by the Kadison–Kaplansky conjecture, so providing a counterexample in the torsion-free case seems to be difficult.
However, every group containing the free product $\Gamma=\mathbb Z/n \ast \mathbb Z$ is non-pr. Indeed, let $g$ denote the generator of $\mathbb Z/n$ and $t$ the generator of $\mathbb Z$, and let $p = \frac1n\sum_{k=0}^{n-1} g^n$ be the projection coming from the torsion part.
The group C*-algebra $C^*_r(\mathbb Z)$ has lots of unitaries which have infinitely many nonzero Fourier coefficients: for instance, $u = \exp(i\cdot(t+t^{-1}))$ (it corresponds via Fourier transform to the function $\exp(2i\cdot\cos\theta)$ on the circle). Let's write its Fourier expansion as $u = \sum_{k\in\mathbb Z} u_kt^k$.
Now, $upu^*$ is clearly a projection in $C^*_r(\Gamma)$. Its Fourier coefficient at $t^k g t^{-\ell}$ is equal to $u_k\overline{u_\ell}/n$, so infinitely many Fourier coefficients are non-zero, and hence in can't belong to $\mathbb C\Gamma$.
I believe, one can relax the freeness condition to something significantly milder (intuitively, one requires an element $t$ of infinite order whose conjugation action on $g$ produces sufficiently many independent elements), but I haven't thought much further.
