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 $f$.
Best Answer
Sorry to refer to my own work, but I think this answers your question directly: http://www.maths.ed.ac.uk/~tl/glasgowpssl/
That link is to a very short note, but I might as well repeat the result here. Let's agree that a "map" of Banach spaces is a map of norm $\leq 1$, and let's also agree that when $X$ and $Y$ are Banach spaces, we equip $X \oplus Y$ with the norm $\| (x, y) \| = (\|x\| + \|y\|)/2$.
Now let $\mathcal{C}$ be the category of triples $(X, \xi, u)$ where $X$ is a Banach space, $\xi$ is a map $X \oplus X \to X$, and $u$ is an element of the closed unit ball of $X$ such that $\xi(u, u) = u$.
Theorem: The initial object of $\mathcal{C}$ is $(L^1[0, 1], \gamma, 1)$ where $1$ is the constant function $1$ and $\gamma$ concatenates two functions then scales the domain by a factor of $1/2$.
Another object of $\mathcal{C}$ is $(\mathbb{R}, \text{mean}, 1)$. The unique map in $\mathcal{C}$ from the initial object to this object is Lebesgue integration, $\int_0^1: L^1[0, 1] \to \mathbb{R}$.
While I'm at it, I'll add another result that isn't in that note (or written up anywhere yet). This characterizes Lebesgue integrability and integration on arbitrary finite measure spaces.
Let $\mathbf{Meas}$ be the category of finite measure spaces and "embeddings" (by which I mean maps that are isomorphisms to their images). Let $\mathbf{Ban}$ be the category of Banach spaces (with maps as above).
Let $\mathcal{D}$ be the category of pairs $(F, u)$, where $F$ is a functor $\mathbf{Meas} \to \mathbf{Ban}$ and $u$ assigns to each measure space $X = (X, \mu)$ an element $u_X \in F(X, \mu)$, subject to two laws: first, $\|u_X\| \leq \mu(X)$, and second, whenever $$ Y \stackrel{i}{\longrightarrow} X \stackrel{j}{\longleftarrow} Z $$ in $\mathbf{Meas}$ with $X = iY \sqcup jZ$ (disjoint union) then $(Fi)u_Y + (Fj)u_Z = u_X$.
Theorem The initial object of $\mathcal{D}$ is $(L^1, I)$, where $I_X \in L^1(X)$ is the constant function $1$.
(In the case of this initial object, the equation "$(Fi)u_Y + (Fj)u_Z = u_X$" says that when $X$ is partitioned into subsets $Y$ and $Z$, the indicator function of $Y$ plus the indicator function of $Z$ is the constant function $1$.)
Another object of $\mathcal{D}$ is $(K, t)$, where $K$ has constant value $\mathbb{R}$ (or $\mathbb{C}$, depending on our choice of ground field) and $t_X = \mu(X)$ for a measure space $X = (X, \mu)$. The unique map in $\mathcal{D}$ from the initial object to this object is integration. To spell that out a bit more: the maps in $\mathcal{D}$ are natural transformations satisfying the obvious condition, and in this case, the $X$-component of the unique map $(L^1, I) \to (K, t)$ is $\int_X: L^1(X) \to \mathbb{R}$.