I should probably have thought a bit more before posting this question: the answer to (2) (and _a fortiori_ to (1)) is _no_. In fact, $C[C^{-1}]$ is the terminal category in the case of interest and so $\iota(C[C^{-1}])$ is contractible.

Recall that a [commutative rig category](https://ncatlab.org/nlab/show/rig+category) $(C,\otimes,I,\oplus,0)$ has symmetric monoidal structures $(C,\oplus,0)$ and $(C,\otimes,I)$, with distributivity $A\otimes 0 = 0$ and $A\otimes(B\oplus C) = (A\otimes B) \oplus (A \otimes C)$ coherently. Alternatively, it is an $E_\infty$-object in symmetric monoidal categories. Let us write $A^\vee$ for a $\otimes$-dual to $A$ and $-A$ for a $\oplus$-dual to $A$.

**Lemma:** Let $(C,\otimes,I,\oplus,0)$ be a commutative rig category. If $0^\vee$ exists, then $0$ is a zero object (i.e. it is both initial and terminal).

**Proof:** By adjunction, we have $C(X,0^\vee) = C(X,0^\vee \otimes I) = C(0 \otimes X, I) = C(0,I)$, naturally in $X$. Taking $X = 0^\vee$ and $1 \in C(0^\vee,0^\vee)$, we obtain a map $X \to 0^\vee$, natural in $X$. Dually, we have a map $0^\vee \to Y$ natural in $Y$, and by composing, we have a map $0_{X,Y}:X \to Y$ natural in $X$ and $Y$. So there is a zero object $\emptyset$ in (the idempotent completion of) $C$. Because $0$ is dualizable, tensoring with it preserves limits and colimits, so we have $\emptyset = \emptyset \otimes 0 = 0$ by distributivity.

**Lemma:** Let $(C,\otimes,I,\oplus,0)$ be a commutative rig category. If $0$ is a zero object and $-A$ exists, then $A=0$.

**Proof:** Because $0$ is a zero object, both the unit and counit maps $A \oplus (-A)^\to_\leftarrow 0$ are zero maps. Then a triangle identity shows that $0_{A,A} = 1_A$, so that $A=0$.

**Corollary:** If $C$ is a commutative rig category with additive duals $-A$ for all objects $A$, and the multiplicative dual $0^\vee$, then $C$ is the terminal category.

**Corollary:** If $C = (Mod(R)^{f.g.proj},\oplus)$, then $C[C^{-1}]$ is the terminal category. Hence $\iota (C[C^{-1}])$ is contractible.

**Proof:** $(C,\otimes,\oplus)$ is a commutative rig category (now $\otimes$ is the usual tensor product of modules). Group completion is a strong monoidal functor on symmetric monoidal categories, so $C[C^{-1}]$ is likewise a commutative rig $\infty$-category, and $C \to C[C^{-1}]$ is a morphism thereof. Therefore, because $0$ is $\otimes$-self-dual in $C$, it is likewise self-dual in $C[C^{-1}]$. Moreover, $C[C^{-1}]$ has additive duals (even better, it has additive inverses). So by the previous Corollary, $C[C^{-1}]$ is the terminal category.