You missed the reduced mass correction to the groundstate energy, which is bigger than the fine structure correction (and of course the Lamb shift). In other words, you missed the motion of the proton. Don't despair - you were right to think that the ionization energy should be very precisely measured and in exact agreement with ($-1$ times) the groundstate energy of hydrogen predicted by the schrodinger equation + fine structure + hyperfine structure + lamb shift + proton radius correction + further dirac equation corrections + higher order lamb shift etc.
I derive this result in more detail here (my most popular answer!). IIRC Griffiths doesn't cover this concept in the Hydrogen chapter because it requires discussion of the concept of a two-particle wavefunction, and it doesn't give rise to any interesting degeneracy splitting like fine structure does. Without fine strucure, the groundstate energy of hydrogen is (in SI units):
$$
-\frac{\mu e^4}{2(4\pi\epsilon_0)^2\hbar^2}
$$
Where $\mu$ is the reduced mass in the center of mass coordinates for the electron and the proton:
$$
\mu=\frac{1}{\frac{1}{m_e}+\frac{1}{m_p}}\approx m_e\left(1-\frac{m_e}{m_p}\right)
$$
Wolfram alpha gives me the value $13.5982873\text{ eV}$. The error is now well in line with fine structure corrections $O(\alpha^4m_ec^2=0.00015\text{ eV})$.
Finally, note that your mentioning of the molecular hydrogen binding energy is totally irrelevant - it's a completely different (and more complicated) quantum mechanical system, and the interaction between the two hydrogens gives a correction that is $O(1)$ (it cannot be thought of as a small correction to the hydrogen wavefunction).
