Ionization energy of S²⁻ vs S
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.