When $S=\mathcal{C}$, it is fairly straightforward to define a symmetric monoidal
structure on the groupoid completion $\mathcal{G}\mathcal{C}:=\mathcal{C}\left[ S^{-1} \right]$. I
couldn't find a reference for this fact, so I'll write it out in this answer.
The goal is to prove the following claim: *the groupoid completion of a symmetric
monoidal category is a symmetric monoidal category*.

Let $\mathcal{C}$ be a small symmetric monoidal category and let $\mathcal{G}\mathcal{C}$ denote the
groupoid completion of $\mathcal{C}$. By the universal property of the groupoid
completion, this defines a functor $\mathcal{G}(-): \mathbf{Cat}\to \mathbf{Gpd}$, to the
category of small groupoids. This functor is left-adjoint:
for any small category $\mathcal{D}$ and any small groupoid $\mathcal{E}$, there
is a natural bijection of sets:
\begin{equation}\label{eq adj}
\mathbf{Gpd}(\mathcal{G}\mathcal{D},\mathcal{E}) \cong \mathbf{Cat}(\mathcal{D}, \mathcal{E}),
\end{equation}
(This is just the universal property.) In fact, this bijection can be enriched to an isomorphism of categories:

**Equation 1**
\begin{equation}
\underline{ \mathbf{Gpd} }(\mathcal{G}\mathcal{D},\mathcal{E}) \cong \underline{ \mathbf{Cat} }(\mathcal{D}, \mathcal{E}).
\end{equation}

By virtue of being a left-adjoint, the functor $\mathcal{G}(-)$ preserves arbitrary colimits.
It turns out that it also preserves finite products:

**Lemma 1:**
For any pair of small categories $\mathcal{C}$ and $\mathcal{D}$, there is a natural isomorphism $\mathcal{G}(\mathcal{C}\times\mathcal{D})\cong\mathcal{G}\mathcal{C}\times\mathcal{G}\mathcal{D}$.

**Proof:**
The proof is purely formal. By the Yoneda lemma, it suffices to exhibit a natural isomorphism between the functors represented by $\mathcal{G}(\mathcal{C}\times\mathcal{D})$ and $\mathcal{G}\mathcal{C}\times\mathcal{G}\mathcal{D}$. For any groupoid $\mathcal{E}$, we have natural isomorphisms
\begin{equation*}
\begin{aligned}
\mathbf{Gpd}(\mathcal{G}\mathcal{C}\times\mathcal{G}\mathcal{D}, \mathcal{E}) & \cong \mathbf{Gpd} (\mathcal{G}\mathcal{C}, \underline{ \mathbf{Gpd} }(\mathcal{G}\mathcal{D}, \mathcal{E}))& \text{by the exponential law for groupoids}\\
& \cong \mathbf{Cat}(\mathcal{C}, \underline{ \mathbf{Gpd} }(\mathcal{G}\mathcal{D}, \mathcal{E})) & \text{by the uni. prop. of $\mathcal{G}\mathcal{C}$}\\
& \cong \mathbf{Cat}(\mathcal{C}, \underline{ \mathbf{Cat} }(\mathcal{D}, \mathcal{E})) & \text{by eq. 1 }\\
& \cong \mathbf{Cat}(\mathcal{C}\times\mathcal{D}, \mathcal{E})& \text{by the exponential law for categories}\\
& \cong \mathbf{Gpd}(\mathcal{G}(\mathcal{C}\times\mathcal{D}), \mathcal{E})& \text{by the uni. prop. of $\mathcal{G}(\mathcal{C}\times\mathcal{D})$}.
\end{aligned}
\end{equation*}

Explicitly, the above isomorphism simply makes morphisms of the form $q_{\mathcal{C}\times\mathcal{D}}(f,g)$ correspond to morphisms of the form $(q_\mathcal{C}(f), q_{\mathcal{D}}(g))$.

Now given a symmetric monoidal category $\mathcal{C}$ with tensor product $\otimes$, we
define a functor $\otimes'$ as the following composite:

Diagram 1

In order to show that $\otimes'$ is part of a symmetric monoidal structure on
$\mathcal{G}\mathcal{C}$, first observe the following.

**Remark:**
Consider the following diagram:

Diagram 2

Here the square commutes by definition of $\mathcal{G}(\otimes)$, and the triangle commutes by definition of $\otimes'$.
Using the explicit description of the isomorphism, we see that commutativity of the above diagram indicates that we have
\begin{equation}
q_\mathcal{C}(f\otimes g) = q_\mathcal{C} (f) \otimes' q_\mathcal{C} (g),
\end{equation}
for any pair of morphisms $f$ and $g$ in $\mathcal{C}$.

Finally we have the following:

**Proposition:**
The groupoid completion of a symmetric monoidal category is a symmetric monoidal category.

**Proof:**
Let $(\mathcal{C}, \otimes, e, \alpha, \lambda, \rho, \gamma)$ be a symmetric monoidal category.
The symmetric monoidal structure on the groupoid completion $\mathcal{G}\mathcal{C}$ is given by $(\otimes', e, q_\mathcal{C}(\alpha), q_\mathcal{C}(\lambda), q_\mathcal{C}(\rho), q_\mathcal{C}(\gamma))$.

The fact that this indeed defines a symmetric monoidal structure follows from
the remark.
To illustrate, we verify the triangle axiom.
Let $a$ and $c$ be objects in $\mathcal{G}\mathcal{C}$ (\ie objects in $\mathcal{C}$).
The triangle axiom for $\mathcal{C}$ states that the following diagram commutes.

Diagram 3

Applying the functor $q_\mathcal{C}$, we see that the following diagram commutes.

Diagram 4

Here we have used that $q_\mathcal{C}(1\otimes\lambda)=1\otimes
q_\mathcal{C}(\lambda)$ and
$q_\mathcal{C}(\rho\otimes1)=q_\mathcal{C}(\rho)\otimes 1$, which follows
from the remark and functoriality of $q_\mathcal{C}$.
This verifies the triangle axiom for $\mathcal{G}\mathcal{C}$, and the other axioms are verified similarly.