Compatibility of two definitions of the projective class group of a group ring

One important ingredient is the following theorem of Swan (Theorem 8.1. in Induced representations and projective modules): If $P$ is any projective f.g. $RG$-module, then $P \otimes_R K$ is free.

With that understood, we can do that following: We actually have a splitting $K_0 (\mathcal P_R ) \cong \mathbb{Z} \oplus \widetilde{K_0 RG}$, where $\mathbb Z$ corresponds to the free modules. (You can define $\widetilde{K_0 RG}$ as the cokernel of the map $K_0 \mathbb{Z} \rightarrow K_0 (\mathcal P_R )$ that sends $\mathbb Z^n$ to $R[G]^n$ and then show that this map splits). The theorem from above guarantees that $\widetilde{K_0 RG}$ is actually the kernel of the homomorphism $K_0 (\mathcal P_R ) \rightarrow K_0 (\mathcal P_K )$ you considered. The group $\widetilde{K_0 RG}$ is called the reduced $K$-theory group of $RG$.

Now the other ingredient is to know that for the group ring $RG$, the locally free class group and the reduced $K$-theory group agree. This is actually non-trivial and is covered for example in Curtis, Reiner Methods of Representation Theory Volume 2, 49.11. Note that what you call the kernel of $\rho_R$ is what they write as $Cl(RG)$.

As a side remark: The reason people considered locally free class groups and reduced algebraic $K$-theory separately is that they are in fact different when applied to an arbitrary order in $KG$. Oftentimes, to compute $K_0(RG)$ it is helpful to go from $RG$ to a maximal order $\Lambda$, compute its $K$-theory (which is often easier) and then check which information got lost along the way.