How does absence of nd orbital in Oxygen affect its valency?
Whether or not oxygen can reach oxidation states higher than $\mathrm{+II}$ is completely independent of the type of orbitals the second shell has. If the orbitals would matter, nitrogen would not be able to reach every single oxidation state between $\mathrm{-III}$ and $\mathrm{+V}$ but would also be restricted to &lt;insert arbitrary number here>.

The actual reason why oxygen’s accessible oxidation states are limited is its very high electronegativity, second only to fluorine among the non-noble gas elements. There is simply no partner that can draw away electrons strongly enough to access high oxidation states. Fluorine itself, while typically being able to access high coordination numbers (higher than oxygen, compare $\ce{XeF6, NF3}$ or $\ce{SF6}$ to $\ce{XeO4, NO3-}$ or $\ce{SO3}$) is not the element of choice for accessing high oxidation states — that is oxygen. But oxygen cannot oxidise itself since it has the same electronegativity. Fluorine itself, as is evident by $\ce{NF5}$ being unknown as of today, would not be able to stabilise an oxidation state of more than $\mathrm{+II}$ on oxygen given our current knowledge.

Furthermore, the statement’s claim that group 16 elements except oxygen only exhibit positive oxidation states is bogus. $\ce{Na2S}$, $\ce{Na2Se}$, $\ce{Na2Te}$ and $\ce{Na2Po}$ are all known and in all of these compounds the chalcogen has an oxidation state of $\mathrm{-II}$.