$(\Leftarrow)$ If $T\circ T^{*}=I$ we had $$\langle x,x \rangle=\langle (T\circ T^{*}) (x),x \rangle=\langle T(x),T(x) \rangle, \forall x\in H.$$ Hence, $\Vert T(x)\Vert=\Vert x \Vert$, $\forall x\in H$.
$(\Rightarrow)$ Suppose that $\Vert T(x)\Vert=\Vert x \Vert$, we had $$\langle T(x),T(x) \rangle=\langle x,x \rangle \Leftrightarrow \langle x,T(x) \rangle=\langle x,T^{*}(x) \rangle \Leftrightarrow \langle x,T(x) -T^{*}(x)\rangle=0,$$ for all $x\in H$. I do not know what to do now...
You are almost there. Note that $$\langle Tx, Tx\rangle = \langle x,x \rangle \iff \langle (T^*T-I)x, x\rangle=0$$
Then invoke the following lemma:
Lemma: If $S$ is an operator on a complex Hilbert space $H$ such that $\langle Sx, x\rangle = 0$ for all $x \in H$, then $S=0$.
Hint: Use the polarization identity. The idea is to use the polarization identity to show that $\langle Sx,y\rangle = 0$ for all $x,y \in H$. Then take $y = Sx$ and conclude that $Sx= 0$ for all $x \in H$, i.e. $S=0.$