I don't think you can construct $T(S)$ as a monoidal category. But you can construct it as a bicategory (with object set $S$).

Let $I(S)$ denote the indiscrete category on $S$. Then a lax 2-functor $I(S) \to M$ (where $M$ is considered as a 1-object bicategory -- more generally $M$ could be an arbitrary bicategory) is the same as an $M$-enriched category with object set $S$ (this goes back to Benabou's original paper on bicategories).

For any bicategory $B$ there is a lax morphism classifier $B'$ such that a weak 2-functor $B' \to C$ is the same as a lax 2-functor $B \to C$ (where $C$ is a bicategory) [actually, I'm not quite certain of this -- if we were talking about strict 2-categories and strict 2-functors rather than weak ones I'd be more certain. You can fit into this framework by replacing $M$ with a monoidal category that is strictly associative]. So $I(S)'$ has the property you're looking for. But it's not exactly a monoidal category -- it's bicategory with many objects.

I think this construction is "well-known". For example, a construction with a similar flavor comes in Gepner and Haugseng's definition of an enriched $\infty$-category: it $S$ is a Kan complex, then let $I(S)$ be the "indiscrete simplicial Kan complex on $S$", with $I(S)_n = S^{n+1}$. A monoidal $\infty$-category $M$ is also sort of simplicial quasicategory, and an $M$-enriched category with object space $S$ is a map of bisimplicial sets $I(S) \to M$ (well, I think they speak in terms of cartesian fibrations over $\Delta$ rather simplicial objects, but these are equivalent by straightening/unstraightening).

I don't think you can construct $T(S)$ as a monoidal category. It's more common to construct $T(S)$ as a bicategory with object set $S$. (see the end for comments on regarding it as a monoidal category)

Let $I(S)$ denote the indiscrete category on $S$. Then a lax 2-functor $I(S) \to M$ (where $M$ is considered as a 1-object bicategory -- more generally $M$ could be an arbitrary bicategory) is the same as an $M$-enriched category with object set $S$ (this goes back to Benabou's original paper on bicategories).

For any bicategory $B$ there is a lax morphism classifier $B'$ such that a weak 2-functor $B' \to C$ is the same as a lax 2-functor $B \to C$ (where $C$ is a bicategory) [actually, I'm not quite certain of this -- if we were talking about strict 2-categories and strict 2-functors rather than weak ones I'd be more certain. You can fit into this framework by replacing $M$ with a monoidal category that is strictly associative]. So $I(S)'$ has the property you're looking for. But it's not exactly a monoidal category -- it's bicategory with many objects.

I think this construction is "well-known". For example, a construction with a similar flavor comes in Gepner and Haugseng's definition of an enriched $\infty$-category: it $S$ is a Kan complex, then let $I(S)$ be the "indiscrete simplicial Kan complex on $S$", with $I(S)_n = S^{n+1}$. A monoidal $\infty$-category $M$ is also sort of simplicial quasicategory, and an $M$-enriched category with object space $S$ is a map of bisimplicial sets $I(S) \to M$ (well, I think they speak in terms of cartesian fibrations over $\Delta$ rather simplicial objects, but these are equivalent by straightening/unstraightening).

If you want $TS$ to be a monoidal category rather than a bicategory, then I think you're in luck because the inclusion from monoidal categories into bicategories should have a left adjoint, just like the inclusion of monoids into category. So just apply this left adjoint to $I(S)'$ above. This adjoint is kind of weird -- you freely add composites for morphisms with non-composable domain and codomain. But in the case of such a simple bicategory as this, I suppose it's not so bad -- and should look basically like you describe.