Question on operator and its adjoint in Hilbert space: $A^{*}A = \text{id}_H$

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Let $\{e_n\}_{n \in \mathbb{N}}$ be a orthonormal sequence in a Hilbert space $H$. Further, let $A : H \to \ell^2$ be a linear operator, and $A^{*} : \ell^2 \to H$ its adjoint, where

$$Ax = \left(\langle x, e_n\rangle\right)_{n \in \mathbb{N}}, \quad\quad x \in H$$

and

$$A^{*}y = \sum_{n=1}^{\infty}y_ne_n, \quad\quad y \in \ell^2.$$

Questions. Is it correct to conclude $A^{*}A = \text{id}_{H}$ if $\{e_n\}_{n \in \mathbb{N}}$ is complete?


I believe so, since for $x \in H$

$$A^{*}Ax = A^{*}\left(\langle x, e_n \rangle\right)_{n \in \mathbb{N}} = \sum_{n=1}^{\infty}\langle x, e_n\rangle e_n$$

which is the orthogonal projection on, $\overline{S} = \overline{\text{span}}\{e_n\}_{n \in \mathbb{N}}$ (usually a theorem in Hilbert space sections). Then, if $\{e_n\}_{n \in \mathbb{N}}$ is complete, the sequence

$$\left(\sum_{n=1}^{k}\langle x, e_n\rangle e_n\right)_{k \in \mathbb{N}} \in S$$

is Cauchy, and converges to $x$. So, $A^{\ast}A = \text{id}_H$. Is this more or less how one shows this?

As a sub-question, does this generally hold for all operators on Hilbert spaces?