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.
Since $k_E \propto \hat{p}^2$ and $\hat{p}$ is Hermitian you may see that this makes $k_E$ positive semidefinite, that is all of its eigenvalues are larger or equal to 0.
In other words when you measure this operator you will always get results which are larger or greater than zero.
This "contradiction" is resolved by the fact that the potential is a function of the coordinate and the momentum (and hence kinetic energy) does not commute with functions of coordinate. This in turn means that you can't simultaneously measure both $k_E$, $E$, and potential.
Best Answer
Bound states correspond to particles that are localised, that is, found in a restricted region of space. In the one dimensional case, you want the probability density to drop to zero rapidly at $x =\pm \infty$ so that the wavefunction is normalisable to $1$.
On the other hand, for scattering states you would have $e^{\pm ikx}$ behaviour with $k$ real, and you cant normalise such states to one on the open real line.
Consider the Schrodinger equation for a one dimensional potential $$ \left( \ \partial^{2}_{x} +k(x)^2 \ \right) \psi(x) =0 \ , $$ with $$k^2(x) = \frac{2m(E-V(x))}{\hbar^2} \ . $$
Lets look at the asymptotic $x \sim \pm \infty$ (that is, large x) solutions of the differential equation. If $E> V(\pm \infty)$ then $k$ is real there and asymptotically the solutions are like the scattering states mentioned above.
For $E<V(\pm\infty)$, $k$ is complex and you can write $k =i\kappa$ to get solutions that behave like $e^{\pm \kappa x}$ asymptotically. These are the bound states (you want to choose the $\pm$ sign for $x = \mp\infty$).
There are probably technicalities to worry about in the above (heuristic) argument, eg if the asymptotic analysis matches smoothly to actual solutions etc. This should work for generic smooth, bounded potentials. You can find details in older books like "Quantum Mechanics" by Messiah or consult texts on properties of second order differential equations.
Here is one interesting exception to the general claim: It IS POSSIBLE to have bound states (normalised to one) embedded among the scattering states (ie when $E> V(\pm \infty)$) if the potential is of the "Von Neumann-Wigner" type: such potentials oscillate indefinitely even as their magnitude vanishes at infinity. An example is discussed in the Quantum Mechanics book by Ballentine. The unusual oscillatory potential traps the particle that you would have expected to escape. As Ballentine says, its a quantum phenomenon without a classical analog in particle mechanics but may be understood in the wave analogy as due to destructive interference.
So the moral probably is this: Generic potentials do indeed have the properties that Griffiths says they do, but one can construct unusual potentials which exploit technical loopholes to evade the folklore theorem.