So far I have reached the fact that is pretty obvious:
$$ \langle v,T^*(u) \rangle = \langle T(v),u \rangle = \langle v,S(u) \rangle = \langle S^*(v),u\rangle $$
thus receiving $T^*(u) = S(u)$ and $T(v) = S^*(v)$.
I know I must prove $T(av+bu) = aT(v) + bT(u)$ and $S(av+bu) = aS(v) + bS(u)$, but I'm very much stuck unfortunately.
This is just a slightly expanded version of the comment I made above. You can make real progress on this question just by lining up what is known. We want to show that $T$ is a linear map, that is$$\label{addition} \tag{$\dagger$} T(u_1+u_2)=T(u_1)+T(u_2),\quad \forall u_1,u_2\in V;$$and $$T(\lambda \cdot v)=\lambda \cdot T(v),\quad \forall \lambda \in \mathbb{R},v \in V.$$(I will assume that $V$ is a real vector space, but this doesn't make a huge difference to how we proceed). Let us first show (\ref{addition}): If $u_1,u_2\in V$ then for all $v \in V$ we have$$\begin{split} \langle T(u_1+u_2),v\rangle & =\langle u_1+u_2,S(v)\rangle =\langle u_1,S(v)\rangle +\langle u_2,S(v)\rangle \\ & =\langle T(u_1),v\rangle +\langle T(u_2),v\rangle =\langle T(u_1)+T(u_2),v\rangle . \end{split}$$Thus we have established that if $s=T(u_1+u_2)$ and $t=T(u_1)+T(u_2)$, then as far as the inner product $(u,v)\mapsto \langle u,v\rangle$ is concerned, $s$ and $t$ are indistinguishable, in that for all $v\in V$ we have $\langle s,v\rangle =\langle t,v\rangle$. But the positivity of the inner product then forces this to mean that $T$ is injective. Indeed if $\langle s,v\rangle =\langle t,s\rangle$ for all $v\in V$, then $\langle s-t,v\rangle =0$ for all $v \in V$ and hence taking $v=s-t$, it follows that $\|s-t\|^2=0$, hence $s=t$ as required.
Checking that $T$ is compatible with scalar multiplication is also easy, since for all $v\in V$ we have$$\langle T(\lambda \cdot u),v\rangle =\langle \lambda \cdot u,S(v)\rangle =\langle u,\lambda \cdot S(v)\rangle =\langle u,S(\lambda\cdot v)\rangle =\langle T(u),\lambda \cdot v\rangle =\langle \lambda \cdot T(u),v\rangle ,$$and so we see that $\lambda \cdot T(u)$ and $T(\lambda\cdot u)$ have $\langle \lambda \cdot T(u),y\rangle =\langle T(\lambda \cdot u),y\rangle$. But then, just as for $s$ and $t$ above, it follows that $T(\lambda \cdot u)=\lambda \cdot T(u)$, and so $T$ is a linear map.