Ionization energy of S²⁻ vs S Which of $\ce{S^2-}$ and $\ce{S}$ has smaller ionization energy?

On one hand, ionization energy of $\ce{S^2-}$ can be predicted to be smaller than that of $\ce{S}$ because it is "easier" to remove and electron from an electron-rich ion, which already experiences a lot of electron-electron repulsion and reduced $Z_\mathrm{eff}$. 

On the other hand, $\ce{S^2-}$ has a stable noble gas electron configuration, and removing an electron would get rid of the noble gas configuration. 

I could not find relevant data for the second electron affinity of sulfur, so I am unable to check the reasoning.
Start by considering that electron affinities and ionization energies are related. Thus $\mathrm{EA}$ of $\ce{S-}$ is equal to $-\mathrm{IE}$ of $\ce{S^2-}$. (Note the convention used here is to report the value of the signed energy change associated with the process (not the absolute value), see the Wikipedia for an explanation if necessary). 

A good place to continue is reading chem.libretexts.org:


  the first electron affinity can be greater than or equal to zero or negative, depending on the electron configuration of the atom. In contrast, the second electron affinity is always positive because the increased electron–electron repulsions in a dianion are far greater than the attraction of the nucleus for the extra electrons.[...] Similarly, the formation of all common dianions (such as $\ce{S^2-}$) or trianions (such as $\ce{P^3−}$) is energetically unfavorable in the gas phase.


This means that the electron affinity of $\ce{S-}$ is positive. Since $\mathrm{IE}(\ce{S^2-}) = -\mathrm{EA}(\ce{S-})$, this means that  $\mathrm{IE}(\ce{S^2-}) &lt; 0$. 

Note in comparison that the first ionization energy of sulfur is $\pu{999.6 kJ/mol}$.

Therefore $\mathrm{IE}(\ce{S^2-}) &lt; \mathrm{IE}(\ce{S})$, with $\mathrm{IE}(\ce{S^2-})$ exothermic.