Naimark Duality Theorem: Let $(\varphi_{i})_{i \in I} \subset \mathcal H$. Then $(\varphi_{i})_{i \in I}$ is a frame iff $\exists$ a Hilbert space $\mathcal K \supset \mathcal H$ and a Riesz Basis $(\psi_{i})_{i \in I}$ for $\mathcal K$ s.t. $\mathcal P_{\mathcal H}\psi_i = \varphi_i$ for all $i \in I$ where $\mathcal P_{\mathcal H}$ is as usual the orthogonal projection of $\mathcal K$ onto $\mathcal H$.
Concerning $\Rightarrow$, the proof starts as follows: Since $(\varphi_{i})_{i \in I}$ is a frame, there is a result that states that the analysis operator $T$ maps $\mathcal H$ bijectively onto a closed subset of $\ell_2(I)$ which we denote by $\mathcal M$. Consider $V: \mathcal M \to \mathcal H, Vc := T^*c$, where $T^*$ the synthesis operator (and adjoint of $T$).
Now let $(e_{i})_{i \in I}$ be the standard orthonormal basis. Since $\ker T^*= \ker P_{\mathcal M} = \mathcal M^⊥$, $V \mathcal P_{\mathcal M} e_i =T^{*} P_{\mathcal M} e_i = T^*e_i = \varphi_i$.
So basically, $e_i$ lies in $\mathcal M$ for all $i \in I$ and $\mathcal M$ is a closed subset of $\ell_2(I)$, then I would follow that $\mathcal M = \ell_2(I)$? That wouldn't make sense.
Let $B:X\to \mathcal{H}$ be any continuous linear operator between Hilbert spaces and let $\mathcal{R}$ be the closure of the range $B(X)$. Then we can write $\mathcal{H}=\mathcal{R}\oplus \mathcal{R}^\perp$. Then if we think about the adjoint, standard techniques tell us that $\ker(B^*)=\mathcal{R}^\perp$.
Now let us write vectors in $\mathcal{H}$ as $(x,y)\in \mathcal{R}\oplus \mathcal{R}^\perp$, where $x\in \mathcal{R}$ and $y\in \mathcal{R}^\perp$. Then $B^*(x,y)=(B^*x,0)$. That is, since $\ker(B)=\mathcal{R}^\perp$, $B^*$ kills the part of $(x,y)$ which is in $\mathcal{R}^\perp$ (which is just $y$), and the value of $B^*$ on $(x,y)$ onlyl depends on $x$.
Now the projection $\mathcal{P}_\mathcal{R}$ on $\mathcal{H}$ has the action $\mathcal{P}_\mathcal{R}(x,y)=(x,0)$. So $\mathcal{P}_\mathcal{R}$ just kills the part of the vector in $\mathcal{R}^\perp$ and keeps the part in $\mathcal{R}$. Then for any $(x,y)$, $$B^*(x,y)=(B^*x,0)$$ and $$B^*\mathcal{P}_\mathcal{R}(x,y)=B^*(x,0)=(B^*x,0).$$ So $B^*=B^*\mathcal{P}_\mathcal{R}$. To explain this in words, if $\mathcal{P}_\mathcal{R}$ is the projection onto the closure of the range of $B$, then $B^*=B^*\mathcal{P}_\mathcal{R}$. This is because $B^*$ only needs to know the part of the vector in $\mathcal{R}$ ($x$ above) and it does not matter what the part of the vector in $\mathcal{R}^\perp$ is ($y$ above) because $B^*$ is just going to send the $y$ to zero anyway. Therefore including the projection $\mathcal{P}_\mathcal{R}$ before $B^*$ only has the effect of keeping the $x$, which is all $B^*$ is acting on anyway. So including the projection $\mathcal{P}_\mathcal{R}$ doesn't change anything, and $B^*=B^*\mathcal{P}_\mathcal{R}$. This is a general phenomenon and doesn't have anything to do with frames. Apply this with $B=T$, $B^*=V$, and $\mathcal{H}=\ell_2(I)$, and $\mathcal{R}=\mathcal{M}$.
To answer your question, everything you have written in your first three paragraphs is correct. But what you have written in the first three paragraphs DOES NOT imply that $e_i\in \mathcal{M}$. By the first part of my answer, $T^*\mathcal{P}_\mathcal{M}x=x$ for ALL $x$, but this does not imply that $x\in \mathcal{M}$. This can also be seen by the two dimensional example I gave in the comments.