Colin's question is a good one. Associativity tells us that \x.1+x and \x.x+1 commute and that's enough -- using the defining universal property of the NNO -- to make them equal: each is a solution to the equations f(0) = 1 and f(s(x)) = s(f(x)). The most conceptual way I can think of to finish the argument is to use that fact that the NNO (in a locos) is the free monoid on a single generator. The opposite-monoid functor is an involution on the category of monoids, hence the opposite monoid, ONN, of NNO is also free. (ONN has the same underlying object as NNO, but with the twisted binary operation.) We need only verify that the unique monoid map g:NNO -> ONN is the identity. Whatever it is, it is satisfies the equations g(0) = 0 and g(x+1) = 1+g(x). And that's enough to force it to be identity.