In answer to the title question: no, you can't always decompose an $L_2$ function in terms of only the bound spectrum of hydrogen.
This is because there are orthogonal functions to all bound states, which naturally represent the free states of the electron. The quickest examples are of course the Coulomb-wave eigenfunctions $|\chi_{E,l,m}⟩$ of the continuous spectrum of hydrogen, and these are not normalizable, but you can take combinations of the form
$$
|\psi⟩=\sum_{l,m}\int f_{lm}(E)|\chi_{E,l,m}⟩\mathrm dE
$$
which do fall in $L_2$, and which are orthogonal to all the bound states since the Coulomb waves $|\chi_{E,l,m}⟩$ are orthogonal to the bound states (since they're eigenfunctions of the same opretaor with different eigenvalues.
This is further clarified in this answer by Arnold Neumaier and in this question; a previous answer to this question states that the continuous Coulomb waves are not formally part of the Hilbert space, but that does not mean that their informal "span" is not.
For more information about these things (including formal proofs of orthogonality and (non)-completeness and so on), I quite like L. D. Faddeev and O. A. Yakubovskii's Lectures on quantum mechanics for mathematics students and L. A. Takhtajan's Quantum mechanics for Mathematicians (with a hat tip to Anatoly Kochubei here for the references).
If you want a concrete example of something that will sit outside of the span of the bound spectrum, simply take a gaussian wavepacket going fast enough,
$$
\psi(\mathbf r)=\left(\frac{2\pi}{\sigma^2}\right)^{3/2}\exp\left(-\frac{(\mathbf r-\mathbf r_0)^2}{2\sigma^2}+ikz\right),
\tag1
$$
project out the bound-state components, and normalize - from physical considerations, you must be left with a nonzero remainder.
As pointed out in Ruslan's comment, the wavefunction in $(1)$ must include a nonzero support outside of the bound-state manifold, because any combination of the form $\sum_{n,l,m}a_{n,l,m}|n,l,m⟩$ will always have a negative expectation value of the energy, whereas the state in $(1)$ will have positive expected energy whenever $k$ and $1/\sigma$ are large enough. That boils down to a routine calculation so I won't perform it here but the core is obvious from physical considerations and if you really care about it then it is a routine procedure to build it into a rigorous argument.
In any case, I'm not sure how any of this answer was unclear, but to be completely explicit:
The Coulomb-wave continuum eigenstates of the hydrogenic hamiltonian are orthogonal to the bound states. (The eigenstates themselves are not in $L^2$, through the usual non-normalizable-state rigamarole, but linear combinations of them can be, and they will remain orthogonal to the bound states.)
The bound states of the hydrogenic hamiltonian are not a complete basis for $L^2(\Bbb R^3)$.
Your question doesn't include any specific examples of sources that claim otherwise, so it's impossible to comment further on why you got the incorrect impression that there is not a consensus about either of the two points above.
Linkedthere). Good question. $\endgroup$