In my notation, $\langle\bullet,\bullet\rangle$ is linear in the first variable and anti-linear in the second variable. Let $Q:H\to H$ be defined by
$$ Qh=h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2$$ for all $h\in H$. Note that $Q$ is hermitian because
\begin{align}\langle Qu,v\rangle &=\big\langle u-\langle u,e_1\rangle e_1-\langle u,e_2\rangle e_2,v\big\rangle\\&=\langle u,v\rangle -\langle u,e_1\rangle \langle e_1,v\rangle -\langle u,e_2\rangle \langle e_2,v\rangle\\&=\langle u,v\rangle -\langle u,e_1\rangle \overline{\langle v,e_1\rangle} -\langle u,e_2\rangle \overline{\langle v,e_2\rangle}\\&=\big\langle u,v-\langle v,e_1\rangle e_1-\langle v,e_2\rangle e_2=\langle u,Qv\rangle.\end{align}
Next, we prove that $Q$ is a projection. That is, $Q^2=Q$. To show this, let $h\in H$ be arbitrary. We have
\begin{align} Q^2h&=Q(Qh)=Q\big(h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2\big)\\&=Qh-\langle h,e_1\rangle Qe_1-\langle h,e_2\rangle Qe_2\\&=Qh-\langle h,e_1\rangle \big(e_1-\langle e_1,e_1\rangle e_1-\langle e_1,e_2\rangle e_2\big) -\langle h,e_2\rangle \big(e_2-\langle e_2,e_1\rangle e_1-\langle e_2,e_2\rangle e_2\big) \\&=Qh-\langle h,e_1\rangle (e_1-e_1-0)-\langle h,e_2\rangle (e_2-0-e_2\rangle\\&=Qh-\langle h,e_1\rangle \cdot 0-\langle h,e_2\rangle\cdot 0= Qh.\end{align}
Now, observe that $Qe_k=e_k$ for $k=3,4,5,\ldots$ but $Qe_1=Qe_2=0$. Therefore, for any $h\in H$, $Qh\perp e_1$ and $Qh\perp e_2$. This is because
$$\langle Qh,e_k\rangle =\langle h,Qe_k\rangle =\langle h,0\rangle =0$$
for $k=1,2$, so $Qh\in \{e_1,e_2\}^\perp =E$. This proves that $\operatorname{im}Q\subseteq E$. The final task to show that for any $h\in E$, $Qh=h$, and this establishes the claim that $\operatorname{im} Q=E$. That is, $Q=P_E$. To see this, we suppose that $h\in E$. Thus, $h\perp e_1$ and $h\perp e_2$, so $\langle h,e_1\rangle=\langle h,e_2\rangle=0$. That is,
$$Qh=h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2=h-0e_1-0e_2=h.$$
I think it is generally true that if $\{e_1,e_2,e_3,\ldots\}$ is an orthonormal basis of a separable Hilbert space $H$ and $P$ is the orthogonal projection onto a closure of the subspace spanned by $\{e_k:k\in A\}$, where $A$ is a subset of $\Bbb N_1$, then $$Ph=\sum_{k\in A}\langle h,e_k\rangle e_k=h-\sum_{k\in \Bbb N_1\setminus A} \langle h,e_k\rangle e_k$$
for all $h\in H$. In other words, $P$ is the projection onto the orthogonal complement of $\{e_k:k\in\Bbb{N}_1\setminus A\}$.
Best Answer
Define the operator $S : H\to H$ by $Sx = \sum_k\langle x,e_k\rangle x_k$. From \begin{align*} \left\|\sum\langle x,e_k\rangle x_k\right\| &\le\left\|\sum\langle x,e_k\rangle (x_k-e_k)\right\| + \left\|\sum\langle x,e_k\rangle e_k\right\|\\ &\le\sum |\langle x,e_k\rangle| \|x_k-e_k\| + \left\|\sum\langle x,e_k\rangle e_k\right\|\\ &\le \left(\sum |\langle x,e_k\rangle|^2\right)^{1/2}\left(\sum \|x_k-e_k\|^2\right)^{1/2} + \left(\sum |\langle x,e_k\rangle|^2\right)^{1/2}\\ &\le 2\left(\sum |\langle x,e_k\rangle|^2\right)^{1/2}, \end{align*} we see that $\sum_{k=1}^n\langle x,e_k\rangle x_k$ is a Cauchy sequence. So, $S$ is well defined. Moreover, if we set $\delta := \left(\sum_k\|x_k-e_k\|^2\right)^{1/2} < 1$, then $$ \|(S-I)x\| = \left\|\sum_k\langle x,e_k\rangle (x_k-e_k)\right\|\le\delta\|x\|. $$ So, $\|S-I\| < 1$ which implies that $S$ is invertible. Now it should be easy for you to show that $(x_k)_k$ is dense in $H$. For this, note that $Se_k = x_k$.