$\mathcal{T}$ is not a Lie group when $n>1$.
Actually, the OP did not say whether he wanted $\mathcal{T}$ to be all possible sequences of compositions of these generating sets, but, if he did, then it is clear that $\mathcal{T}$ is not a Lie group, in the sense that it is not defined as the set of solutions of some system of PDE for transformations of $\mathbb{R}^n$. For one thing, the group that they generate would properly contain the conformal group $\mathrm{O}(n{+}1,1)$ acting on $S^n$, which is known to be a maximal Lie group, i.e., there is no group (in Lie's sense) between the conformal group and the full diffeomorphism group. (NB: The group of analytic diffeomorphisms of $S^n$ is not a subgroup of the full diffeomorphims in Lie's sense because it is not defined as the set of solutions of some system of PDE.)
In particular, no group $G$ that contains $\mathcal{T}$ can preserve any geometric structures of the kind the OP mentions because this would define a PDE that $G$ satisfies.
(By the way, note that $\mathcal{T}$, as the OP defined it, does not consist of smooth transformations of $S^n$ only when $n>1$, since the non-conformal affine transformations do not extend smoothly to $\infty$ except when $n=1$.)