How can I complete the final step? I cannot build any connection between (2) and (3).
You need to use the completeness of the eigenfunctions
$$ \sum_i \psi_i(\vec r_1)\psi_i^*(\vec r_2) = \delta(\vec r_1 - \vec r_2) $$
After plugging in the above completeness relationship on the RHS and your definition of the Green's function on the LHS, the resulting equation is:
$$ \sum_i \psi_i^*(\vec r_2)\frac{1}{(\omega - E_i)}\left \{(-\omega - h(\vec r_1))\psi_i(\vec r_1) - \int d^3r_3 \Sigma(\vec r_1,\vec r_3)\psi_i(\vec r_3)\right\} =\sum_i \psi_i(\vec r_1)\psi_i^*(\vec r_2) $$$$ \sum_i \psi_i^*(\vec r_2)\frac{1}{(\omega - E_i)}\left \{(\omega - h(\vec r_1))\psi_i(\vec r_1) - \int d^3r_3 \Sigma(\vec r_1,\vec r_3)\psi_i(\vec r_3)\right\} =\sum_i \psi_i(\vec r_1)\psi_i^*(\vec r_2) $$
Note, In the above equation I have corrected an error in the Fourier transform of the LHS in your original statement of the problem. Your $(-\omega -h(\vec r))$ has been changed to $(\omega - h(\vec r))$. This is because the $i\partial/\partial t$ transforms to $\omega$ not $-\omega$.
The coefficients of $\psi_i^*(\vec r_2)$ have to be equal by completeness so: $$ \frac{1}{(\omega - E_i)}\left \{(-\omega - h(\vec r_1))\psi_i(\vec r_1) - \int d^3r_3 \Sigma(\vec r_1,\vec r_3)\psi_i(\vec r_3)\right\} =\psi_i(\vec r_1) $$$$ \frac{1}{(\omega - E_i)}\left \{(\omega - h(\vec r_1))\psi_i(\vec r_1) - \int d^3r_3 \Sigma(\vec r_1,\vec r_3)\psi_i(\vec r_3)\right\} =\psi_i(\vec r_1) $$ Or $$ (-\omega - h(\vec r_1))\psi_i(\vec r_1) - \int d^3r_3 \Sigma(\vec r_1,\vec r_3)\psi_i(\vec r_3) =\psi_i(\vec r_1)(\omega - E_i) $$$$ (\omega - h(\vec r_1))\psi_i(\vec r_1) - \int d^3r_3 \Sigma(\vec r_1,\vec r_3)\psi_i(\vec r_3) =\psi_i(\vec r_1)(\omega - E_i) $$
Cancel the $\omega \psi_i(\vec r_1)$ from both sides and then multiply both sides by $-1$ to get: $$ h(\vec r_1)\psi_i(\vec r_1) + \int d^3r_3 \Sigma(\vec r_1,\vec r_3)\psi_i(\vec r_3) =\psi_i(\vec r_1)E_i $$