The carbon on the right hand side of the double bond did start at an oxidation state of 0 then stayed 0 in the radical intermediate and finally became +1 in the final product. So, it was oxidized. Its a pain to view organic mechanisms that way, but it can sometimes help.
On an olefin you count one of the carbons as in the alcohol oxidation state and one in the alkane oxidation state, right? I guess I don't see the oxidation of the carbon on the right of the alkene. If you count the one on the left as in the alcohol oxidation state, then it stays the same and the one on the right stays in the alkane oxidation state.
Maybe it's just the way crazy organic chemists glaze over formal oxidation states....
To me it seems that the oxidation really happens when the N-S bond breaks since the S takes an electron from N. The hydride abstraction at the end is a reduction, if anything, right?