Localization of a symmetric monoidal category is monoidal when the morphisms being inverted are closed under tensor product In the answer to question Localization of symmetric monoidal category, it was mentioned that '' Assuming that the tensor product of two morphisms in $S$ is again in $S$, the localised category should inherit a symmetric monoidal structure, just by the universal property.''
So I want to know by which universal property we can show that $\mathcal{M}[S^{-1}]$ inherit a symmetric monoidal structure?
Since I cannot comment on the original answer, I posted this as a new (stupid) question.
 A: A reference regarding the monoidal structure on $M[S^{-1}]$ is Brian Day's Note on monoidal localisations. It also talks about the enriched setting, and about monoidal completion. It includes a proof of Neil Strickland's point about requiring $S$ to be closed under $\otimes$, from the other MO thread you linked to.
A: 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.
