Suppose that
$$\Psi = ke^{ar_1}e^{ar_2}$$
and it is known that
$$ \left(E_0 + \frac{1}{\lvert r_1 - r_2 \rvert}\right)\Psi = E\Psi$$
for some constants $E, E_0$.
Assuming that $E$ can be expressed as a perturbation series
$$E = E_0 + E_1 + E_2 + ...$$
we get
$$ \left(E_0 + \frac{1}{\lvert r_1 - r_2 \rvert}\right)\Psi = (E_0 + E_1 + E_2 + ...)\Psi$$
and so
$$ \left(\frac{1}{\lvert r_1 - r_2 \rvert}\right)\Psi = (E_1 + E_2 + ...)\Psi$$
Taking the expectation value of $\frac{1}{\lvert r_1 - r_2 \rvert}$:
$$ \langle \frac{1}{\lvert r_1 - r_2 \rvert} \rangle = \int \bar \Psi \frac{1}{\lvert r_1 - r_2 \rvert} \Psi d\tau = \frac{20\pi k}{a^5}$$
Thus
$$ \frac{20\pi k}{a^5} = \langle E_1 + E_2 + E_3 + E_4 + ...\rangle $$ $$= E_1 + O(E_2 + E_3 + E_4 + ...)$$
Therefore we can take $E_1 = \frac{20\pi k}{a^5}$.
Knowing $E_1$ from $$ \left(\frac{1}{\lvert r_1 - r_2 \rvert}\right)\Psi = (E_1 + E_2 + E_3 + ...)\Psi$$
we get
$$ \left(\frac{1}{\lvert r_1 - r_2 \rvert} - E_1\right)\Psi = (E_2 + E_3 + ...)\Psi$$
Taking expectation value of $\frac{1}{\lvert r_1 - r_2 \rvert} - E_1$:
$$ E_2 = \langle \frac{1}{\lvert r_1 - r_2 \rvert} - E_1 \rangle = \int \bar \Psi \left( \frac{1}{\lvert r_1 - r_2 \rvert} - E_1 \right) \Psi d\tau = ?$$
Can $\int \bar \Psi \left( \frac{1}{\lvert r_1 - r_2 \rvert} - E_1 \right) \Psi d\tau$ be computed simply from our knowing $\int \bar \Psi \frac{1}{\lvert r_1 - r_2 \rvert} \Psi d\tau= \frac{20\pi k}{a^5}$?
Suppose
$$\Psi = ke^{-ar_1}e^{-ar_2}$$
Then
$$E_2 = \int \bar \Psi \left( \frac{1}{\lvert r_1 - r_2 \lvert} - E_1 \right)\Psi d\tau$$ $$=\int k^2 e^{-2ar_1}e^{-2ar_2}\left( \frac{1}{\lvert r_1 - r_2 \lvert} - E_1 \right) d\tau$$ $$=k^2\int \left( \frac{e^{-2ar_1}e^{-2ar_2}}{\lvert r_1 - r_2 \lvert} \right) d\tau - k^2E_1 \int e^{-2ar_1}e^{-2ar_2} d\tau $$ $$=\left( \frac{20\pi^2k^2}{a^5} \right) - \left( \frac{20\pi^2k^2}{a^5} \right)\int e^{-2ar_1}e^{-2ar_2} d\tau$$
Evaluating $\int e^{-2ar_1}e^{-2ar_2} d\tau$:
$$\color{red}{\int_{\phi_1=0}^{2\pi}} \color{orange}{\int_{\theta_1=0}^\pi} \int_{r_1=0}^\infty e^{-2ar_1}r_1^2dr_1\color{orange}{\sin\theta_1d\theta_1}\color{red}{d\phi_1\int_{\phi_2=0}^{2\pi}} \color{orange}{\int_{\theta_2=0}^\pi} \int_{r_2=0}^\infty e^{-2ar_2}r_2^2dr_2\color{orange}{\sin\theta_2d\theta_2}\color{red}{d\phi_2}$$ $$=\color{red}{(2\pi)^2}\color{orange}{(2)^2}\int_{r_1=0}^\infty r_1^2e^{-2ar_1}dr_1\int_{r_2=0}^\infty r_2^2e^{-2ar_2}dr_2$$ since $$\color{red}{\int_0^{2\pi} d\phi = 2\pi}$$ $$\color{orange}{\int_0^\pi \sin\theta d\theta = 2}$$ Using integration by parts $$= \color{red}{(2\pi)^2}\color{orange}{(2)^2} \color{blue}{\left[ -\frac{\bar r_1^2}{2}e^{-2\bar r_1} - \frac{\bar r_1}{2}e^{-2 \bar r_1} - \frac{e^{-2\bar r_1}}{4} \right]^\infty_0}\color{blue}{\left[ -\frac{\bar r_2^2}{2}e^{-2\bar r_2} - \frac{\bar r_2}{2}e^{-2 \bar r_2} - \frac{e^{-2\bar r_1}}{4}\right]^\infty_0}$$ $$= \color{red}{(2\pi)^2}\color{orange}{(2)^2} \color{blue}{\left( \frac{1}{4} \right)}\color{blue}{\left( \frac{1}{4} \right)}$$ since $$\color{blue}{\int_{r=0}^\infty r^2e^{-2ar}dr = \frac{1}{4}}$$ Therefore,
$$\boxed{\int e^{-2ar_1}e^{-2ar_2} d\tau=\pi^2}$$
Combining $$E_2=\left( \frac{20\pi^2k^2}{a^5} \right) - \left( \frac{20\pi^2k^2}{a^5} \right)\color{purple}{\int e^{-2ar_1}e^{-2ar_2} d\tau}$$ $$=\left( \frac{20\pi^2k^2}{a^5} \right) - \left( \frac{20\pi^2k^2}{a^5} \right)\color{purple}{\pi^2}$$
Thus, $E$ to the third approximation is $$E = E_0 + E_1 + E_2 + ... $$ $$= E_0 + \frac{20\pi^2}{a^5}k + \left( 1-\pi^2 \right)\left( \frac{20\pi^2}{a^5}\right)k^2+...$$
Suppose $$\Psi_1 = ke^{-ar_1}e^{-br_2}$$ $$\Psi_2 = ke^{-br_1}e^{-ar_2}$$
By symmetry,
$$E_0 (ke^{-ar_1}e^{-br_2})= E_0 (ke^{-br_1}e^{-ar_2})$$ $$E_0 \Psi_1 = E_0 \Psi_2$$
Hence $\Psi_1, \Psi_2$ are indistinguishable by their $E_0$.
Suppose an asymmetry can be introduced such that
$$(E_0 + \frac{1}{r_1})\Psi = E\Psi$$
Then
$$ E_1 = \langle \frac{1}{r_1} \rangle = \int \bar \Psi \frac{1}{r_1} \Psi d\tau$$
$$ = k^2 \int r_1 e^{-2ar_1} d\tau_1 \int \int \int r_2^2 e^{-2br_2} d\tau_2$$
$$ = (2\pi)^2(2)^2k^2\left[\frac{1}{-2a}\right]^2 \left[\frac{1}{-2b}\right]^3$$
$$ = -2^4\pi^2k^2\left[\frac{1}{2^2a^2}\right] \left[\frac{1}{2^3b^3}\right]$$
$$ = -\left[\frac{\pi^2k^2}{2a^2b^3}\right]$$
Due to this asymmetry, $E_1$ is now distinguishable between $\Psi_1, \Psi_2$, since for $b \neq a$: $$-\left[\frac{\pi^2k^2}{2a^2b^3}\right] \neq -\left[\frac{\pi^2k^2}{2b^2a^3}\right]$$