The answer is no in general: let $R= \mathbb Z$ and $E= \mathbb Z/p$. Then $[E] = 0$ because of the co/fiber sequence $\mathbb Z\to \mathbb Z\to \mathbb Z/p$.

However, by the theorem of the heart, $K_0(C)= K_0(\mathbb Z/p)$ (of course $C$ is not equivalent to $Perf(\mathbb Z/p)$ but it is still true on $K$-theory) which is clearly nonzero.

Here is one situation where something like this holds: if $C$ is generated without retracts by $E$, and the thick closure of $C$ is the whole of $Perf(R)$. In this case, $C$ is a so-called ''dense'' subcategory and Thomason proved that $K_0(C)$ injects into $K_0(Perf(R))$ as the subgroup generated by $[E]=0$.

The answer is no in general: let $R= \mathbb Z$ and $E= \mathbb Z/p$. Then $[E] = 0$ because of the co/fiber sequence $\mathbb Z\to \mathbb Z\to \mathbb Z/p$.

However, by the theorem of the heart, $K_0(C)= K_0(\mathbb Z/p)$ (of course $C$ is not equivalent to $\operatorname{Perf}(\mathbb Z/p)$ but it is still true on $K$-theory) which is clearly nonzero.

Here is one situation where something like this holds: if $C$ is generated without retracts by $E$, and the thick closure of $C$ is the whole of $\operatorname{Perf}(R)$. In this case, $C$ is a so-called ''dense'' subcategory and Thomason proved that $K_0(C)$ injects into $K_0(\operatorname{Perf}(R))$ as the subgroup generated by $[E]=0$.