The scattering states must be included in the perturbative calculations if the result is to be highly accurate. In particular, it is not justified to ignore the continuous spectrum at energies close to the dissociation threshold.
The Hilbert space in the position representation is the space of square integrable functions on $R^3\setminus\{0\}$ with respect to the inner product
$$\langle\phi|\psi\rangle:=\int \frac{dx}{|x|}\phi(x)^*\psi(x).$$
The bound states alone are not dense in this space.
For thorough treatments of the hydrogen spectrum see the books
G R. Gilmore,
Lie Groups, Lie Algebras & Some of Their Applications,
Wiley 1974, Dover, 2002. pp. 427-430
and
A.O. Barut and R. Raczka,
Theory of group representations and applications, 2nd. ed.,
Warzawa 1980. Chapter 12.2.
As you rightly note, all of the doubly excited states of helium are above the single-ionization threshold, and this means that they are auto-ionizing: they have enough energy to slingshot one of the electrons into the continuum and, given half the chance, will proceed to do so.
The only reason that this doesn't happen in the naive theory is that the electrons simply don't 'know' about the existence of the other guy to an extent that's sufficient for the transfer of actual quantum mechanical energy between them. That is, you can add self-consistent mean-value-of-the-density contributions to the (now nonlinear and nonlocal) hamiltonian for each electron, but you are still, by the simple fact that you're solving single-particle Schrödinger equations, explicitly forbidding the kinds of configuration interactions that make autoionizing states spring to life.
As such, any sort of entangling interaction between the two electrons will act as a link between your (formerly) bound states and the continuum, and it will force the eigenstates to be a linear combination of both: that is, the bound state becomes a non-square-integrable resonance embedded in the continuum, with an imaginary part of the eigenvalue which requires the state to decay over time as population goes away towards infinity.
The example you mentioned, $\mathrm{sech}(|\mathbf r_1-\mathbf r_2|)$, is a good candidate, as would be a softened Coulomb potential, but you could also use e.g. a contact interaction, $\delta(\mathbf r_1-\mathbf r_2)$, and get much the same results. (In fact, those would be pretty interesting to see, but I'm not aware of any references and it sounds like a hard thing to search for.)
If you want the abstract theory, there's probably no place better than the original Fano paper on Fano resonances,
U. Fano. Effects of configuration interaction on intensities and phase shifts. Phys. Rev. 124, 1866 (1961), eprints.
It has aged a bit, but it is still pretty readable, and the depth of its arguments makes up for its slightly outdated notation. In particular, it is very general regarding what sorts of particle-particle coupling would give rise to the autoionizing character of those states: all you need is a nonzero entangling matrix element, and it will happen. (Even better, so long as that matrix element is reasonably flat along the relevant bit of continuum, you also get the Fano line profile as a universal characteristic, with only the $q$ parameter tuning the shape.)
Best Answer
From the Virial Theorem we can say the total energy of the atom is propotional to the potential energy of the atom. The potential energy is given by a coulomb potential and so is it will be roughly proportional to $\frac{1}{\langle r \rangle}$ where $\langle r \rangle$ is the mean radius of the electron's orbital. For a hydrogen atom the energy $E\propto \frac{1}{n^2}$, so we would expect \begin{align}\frac{1}{\langle r \rangle} & \propto \frac{1}{n^2}\\ \langle r \rangle &\propto n^2\end{align}
Unfortunately this does not help you much in storing your infinite amount of information in a single atom. In order to get an estimate of $\langle r \rangle$ you need to make many measurements of the position of the electron (especially if it is in a very spread out distribution such as for a high $n$ state) each of these measurements will collapse the wavefunction and you will have to prepare the atom into its initial state all over again before making the next measurement... but that was exactly what you were trying to ascertain by measuring the electron! To determine the state of your atom you need a set of quantities that can be measured simultaneously and uniquely determine which state your atom was in first time, which for a hydrogen atom essentially means you want the energy and the angular momentum. Now the angular momentum shouldn't give you too much trouble; it is nicely quantised into integer multiples of $\hbar$. The energy however will give you a problem. As $E \propto -n^{-2}$, for large $n$, the separation of energy levels goes like \begin{align} \frac{1}{n^2} - \frac{1}{(n+1)^2} &= \frac{1}{n^2}\left[ 1 - (1+ \frac{1}{n})^{-2}\right]\\ &\approx \frac{2}{n^3}\end{align} That will get very small very fast and how accurately you can measure that separation will limit how much information you can practically store.