You don't necessarily need to have your element binding with oxygen to be oxidized, per se. Unless it is, of course, specified that we're talking about a reaction with oxygen. So, don't misguide yourself on focusing on the oxygen here :) After all, if we, for instance, think about burning of potassium, it is being oxidized with oxygen, however, it still only makes $\ce{K+}$ and not $\ce{K^2+}$!
The trick in this question is to look for the values, or a magnitude, of those ionization energies in question. To remove the first electron, you need to provide $\pu{7 eV}$, to remove the second one, you need to provide $\pu{12.5 eV}$. So the difference between 7 and 12.5 is not that big if we compare it with the difference between $\pu{12.5 eV}$ and $\pu{42.5 eV}$ that is required to remove the third electron. If explaining it in simple terms, the first to energy "bumps" are easy to "climb", while the last one is way high and you really need to provide a sufficient amount of energy to get that high. Thus, removing the first two electrons is relatively easy, while removing the third one is not.
This gives you the most stable oxidation state for this element to be +2. If we perform a reaction of this element (which looks like $\ce{Sr}$) with any other electronegative element, say chlorine, it would still have a charge of +2.
