I'll try to bend David White's answer towards the actual situation of your question. The outcome is somewhat clumsy and it totally looks like model structures can be eliminated from it, but anyway:
Assume your category C is closed monoidal and locally presentable. Then it is a monoidal model category with cofibrations and fibrations all morphisms and weak equivalences the isomorphisms. This model category is cofibrantly generated: One can take the identity of the initial object as generating trivial cofibration and the set of all morphisms between the objects $G$ from some generating set as generating cofibrations.
This model category then satisfies the hypotheses of Barwick's Thm. 4.46 in this article for a Bousfield localization at the one element set containing $f$. The homotopy category for the localized model structure has the universal property you want and lives in the same universe. You have an adjunction between the homotopy category of the original model structure, which is the category itself, and the localization.
This adjunction is a reflection to an orthogonal subcategory as in Adamek/Rosicky, 1.35-1.38, namely to the full subcategory of all objects from whose point of view $f$ "was already an isomorphism" (i.e. $f$-orthogonal objects; precise definition via a unique-lifting-condition). This is much like in your example (but with the condition on the twist removed and without the domain of f having to be special). If you chase through the proofs, you also get an expression of the reflection functor as a colimit via the small object argument, resp. via Adamek/Rosicky's proof...
Barwick's Prop. 4.47 gives then a criterion for the homotopy category to be closed monoidal again: It suffices that any object $X$ which satisfies the unique right lifting condition with respect to $f$ also satisfies it with respect to $f \otimes G$ for every generating object $G$ (remember the category was locally presentable now) i.e. if $f$ induces an iso $Hom(f,X)$ then $Hom(f \otimes G,X)$ is an iso, too, for every generating object $G$.
edit: Sorry, I am no longer sure that the homotopy category of the Bousfield localization is in fact the localization along $f$ in the sense you asked for: When you localize with respect to an arrow you automatically invert together with it a bunch of other arrows. When you do plain category theory it is somewhat uncontrollable what those other arrows are, it seems to me. When you do Bousfield localization these other arrows are those having the left lifting property with respect to the $f$-local objects. Now I don't see a reason why the class of additionally inverted arrows should be the same in both cases. What Bousfield localization as sketched here probably yields, is the universal colimit preserving functor which inverts $f$.
First, having seen the edited version of Martin's question, let's quickly dispose of the construction of the free symmetric monoidal category generated by a category $C$. Objects are tuples $(x_1, \ldots, x_n)$ of objects of $C$. Morphisms are labeled permutations, where permutations are conveniently visualized as string diagrams, each string being labeled by a morphism in $C$. To compose such labeled diagrams, just compose the string diagrams, composing the labels of strings in $C$ along the way.
The free symmetric monoidal category $Sym(M)$ on a monoidal category $M$ is formed from the free symmetric monoidal category $S (U M)$ on the underlying category $U M$ by adjoining isomorphisms $\phi_{x_1, \ldots, x_n}: (x_1, \ldots, x_n) \to x_1 \otimes \ldots \otimes x_n$, where the $x_i$ are objects of $M$, the expression $(x_1, \ldots, x_n)$ is the formal monoidal product in $S(UM)$, and $x_1 \otimes \ldots \otimes x_n$ is the tensor product in $M$.
More precisely, define the objects of $Sym(M)$ to be tuples $(x_1, \ldots, x_n)$ of objects of $M$. Define morphisms $(x_1, \ldots, x_n) \to (y_1, \ldots, y_m)$ to be equivalence classes of pairs $(p, f)$ where $p$ is a permutation on $n$ elements, and $f$ is an $M$-labeled planar forest of $m$ rooted trees. (You should think here of the free multicategory generated by a category.) Formally, a planar forest can be described as a functor $[n]^{op} \to \Delta$ where $n$ is the category $1 \leq \ldots \leq n$ and $\Delta$ is the category of finite (possibly empty) ordinals. Under the obvious way of drawing such forests, the ordered list of leaves of the forest is labeled by $(x_{p(1)}, \ldots, x_{p(n)})$, and the ordered list of roots by $(y_1, \ldots, y_m)$. Edges are labeled by objects of $M$, and each internal node with $k$ inputs labeled (in order) by $m_1, \ldots, m_k$ and output $m$ is labeled by a morphism $f: m_1 \otimes \ldots \otimes m_k \to m$.
There is an evident way, using the monoidal structure of $M$, of evaluating such a labeled forest $f$ as a morphism $ev(f): x_{p(1)} \otimes \ldots \otimes x_{p(n)} \to y_1 \otimes \ldots \otimes y_m$ in $M$. We consider two arrows $(p, f)$ and $(p', f')$ to be equivalent if $p = p'$ as permutations and $ev(f) = ev(f')$.
Now we define composition of pairs. The main idea is to rewrite the composition of a forest followed by a permutation,
$$(x_1, \ldots, x_n) \stackrel{(1, f)}{\to} (y_1, \ldots, y_m) \stackrel{(q, 1)}{\to} (y_{q(1)}, \ldots, y_{q(m)}),$$
into a form $(p, f')$ forced by the naturality requirement of the symmetry isomorphism. Namely, if $\bar{x}_i$ is the tuple of leaves of the tree whose root is $y_i$, then we have a block permutation $bl(q)$ taking $(\bar{x}_1, \ldots, \bar{x}_m)$ to $(\bar{x}_{q(1)}, \ldots, \bar{x}_{q(m)})$. The permutation $q$ can also be applied to the $m$ trees of the forest by reordering the trees, yielding a new forest $\mathrm{perm}_q(f)$, and the composition $(q, 1) \circ (1, f)$ is defined to be $(bl(q), \mathrm{perm}_q(f))$. Then, if we have $(p, f): (x_1, \ldots, x_n) \to (y_1, \ldots, y_m)$ and $(q, g): (y_1, \ldots, y_m) \to (z_1, \ldots, z_p)$, we define their composite to be
$$(bl(q) \circ p, g \circ \mathrm{perm}_q(f))$$
where the first component is by composing permutations and the second is by the usual way of composing forests by plugging in roots of one planar forest of trees into leaves of another.
Remaining details that all this works will be left to the reader. I will remark that the key rewriting procedure above is an instance of a kind of distributive law; there are some more details to this effect in some notes on my nLab web; see here.
Best Answer
One needs that $C$ is cocomplete and that $\otimes$ preserves colimits in each variable (and then $\mathrm{Mod}_C(A)$ will have the corresponding properties). More precisely, you only need that $C$ has reflexive coequalizers and that $\otimes$ preserves them in each variable. This has been known for decades, but the first clear write-up of this, at least I know of, is Florian Marty's thesis. You can also find a discussion on this in my thesis, Section 4.1 (and Chapter 6 for the issue on reflexive coequalizers). See also MO/114457 for a discussion of the internal homs by Todd Trimble.
Edit. Here are some other references, which even discuss the case where $- \otimes A$ is replaced by a (suitable) symmetric monoidal monad:
If I recall correctly, some of these references restrict the monads in such a way that they are of the form $- \otimes A$ anyway.