The distinction to be made here is that, for the quantum harmonic oscillator system, there are no unbound states, only bound states thus, there is no benefit to insisting the states have negative energy, no reason to 'subtract infinity' in order to zero the potential at infinity.
However, in systems that permit both bound and unbound states, it is reasonable to zero the potential at infinity for the same reason that we do this classically.
For example, in the classical central force problem, there is a state in which particle can 'escape to infinity' where it will have zero kinetic energy (more precisely, the kinetic energy of the particle asymptotically approaches zero). If we set the potential energy to be zero at infinity, then the total energy 'at infinity' is zero. Thus, the particle with zero total energy 'sits on the boundary' between those particles with not enough energy to 'reach' infinity and those that do.
But, for the classical harmonic oscillator potential, no particle can escape to infinity. The kinetic energy of the particle will periodically and instantaneously be zero. In this case, it is reasonable that the state where the total energy is always equal to the potential energy (the state where the kinetic energy is always zero) be the zero total energy state; all other states having positive total energy.
So the conclusion is that nothing really exists according to Quantum
Mechanics... which can't be right, surely?
That's not remotely the correct conclusion to draw. One might conclude instead that
(1) The conception of bound state must be modified in the passage from classical mechanics to quantum mechanics and
(2) the physical (normalizable) unbound states are not eigenstates of the Hamiltonian, i.e., the physical unbound states are not states of definite energy but are, instead, a distribution of energy eigenstates, e.g., a wavepacket.
Best Answer
This definition of bound and scattering states is not quite correct, although it holds for many potentials. There are counterexamples to this fact that have roots in a paper by von Neumann and Wigner. One is the spherical potential $$V(r)=32\sin r \frac{\sin r-(1+r)\cos r}{\left(2+2r-\sin 2r\right)^2}$$ It is not hard to check that $V(r)$ is a bounded continuous function that vanishes at infinity. Even though, the function $$\psi(x)=\frac{2\sin r}{r\left(2+2r-\sin 2r\right)}$$ is an eigenfunction of $H=-\Delta + V$, with eigenvalue $1>0$.
So this is an example of a bound state for which these conditions do not hold. The precise definition of bound states is more subtle and is given by the elements of $$\mathcal{H}_{\text{bound}}(H)=\left\{\psi(x,0) \in L^2(\Bbb{R}^n):\lim_{r\to\infty}\sup_{t\in \Bbb{R}} \int_{\Bbb{R^n}\backslash B(0;r)}|\psi(x,t)|^2dx=0\right\},$$ where $\psi(x,t)=e^{-itH}\psi(x,0)$, that is, the states that are localized in space for any time $t$. It's always true that $\mathcal{H}_{\text{bound}}(H) \supset \mathcal{H}_{\text{p}}(H)$, the closure of the set of linear combinations of eigenvectors of $H$, and for some potentials the equality holds.
For the delta-function potential, the realization of a self-adjoint operator that has the right properties is not so simple, but can be done in some ways. But as Griffiths discuss, the change from a negative to positive delta potential kills the bound state, since its only eigenfunction is not normalizable anymore and all states are scattering states.