Prove that V is an identity operator

Fix any basis $b$ and consider the matrix $A=V_b$. Then you know that there exist invertible matrices $S$ and $T$ such that $S^{-1}AS$ is unitary and $T^{-1}AT$ is positive definite.

Since similar matrices share eigenvalues, you see that the eigenvalues of $A$ have modulus $1$ (because $S^{-1}AS$ is unitary) and positive real (because $T^{-1}AT$ is positive definite).

This only leaves tha possibility that $A$ has the single eigenvalue $1$.

Now use that positive definite matrices are diagonalizable.

We have that $V$ is unitary and that $V$ is self-adjoint and positive definite. Hence

$$V^{-1}= V^*$$

and

$$V=V^*.$$

This gives $V^2=I$ thus

$$(V-I)(V+I)=0.$$

Since $V$ is positive definite, $-1$ is not an eigenvalue of $V$, therefore $V+I$ is invertible. This gives

$$V=I.$$

$V_f$ has the same spectrum as $V_e$ because both matrices are matrix representations of the same linear operator $V$ in some bases. Hence the eigenvalues of $V_f$ are positive real numbers (because $V_f$ is positive definite) with unit moduli (because $V_e$ is unitary), i.e. all eigenvalues of $V_f$ are equal to $1$. Hence $V_f=I$ and $V$ is the identity operator.