Here's a straightforward observation. If you assume $NP \neq coNP$, then it is pretty easy to see there are $NP$ optimization problems which do not even have good nondeterministic approximation algorithms, in some sense.
For example, the PCP theorem says that you can translate SAT into the problem of distinguishing whether $1-\varepsilon$ of the clauses are satisfied and all of the clauses are satisfied, for some $\varepsilon > 0$. Suppose there is a nondeterministic algorithm which can distinguish between these two cases, in the sense that the nondeterministic algorithm can report in each computation path either "all satisfied" or "at most $1-\varepsilon$", and it says "at most $1-\varepsilon$" in some path if at most $1-\varepsilon$ can be satisfied, otherwise it says "all satisfied" in every computation path if all equations can be satisfied. This is enough to decide SAT in $coNP$, so $NP=coNP$. It seems clear that the existence of such a nondeterministic algorithm has no bearing on whether $P = NP$.
It's quite plausible that a more "natural" scenario exists: an optimization problem which is hard to approximate in deterministic polynomial time under $NP \neq coNP$ but not known to be hard under $P \neq NP$. (This is probably what you really wanted to ask.) Many hardness of approximation results are first proven under some stronger assumption (e.g. $NP$ not in subexponential time, or $NP$ not in $BPP$). In some cases, later improvements weaken the necessary assumption, sometimes down to $P \neq NP$. So there is hope that there's a slightly more satisfactory answer to your question than this one. It is hard to wonder how there could be a problem that cannot be proved hard to approximate in deterministic polytime under $P \neq NP$, but it can be proved hard under $NP \neq coNP$. That would mean that $NP \neq coNP$ tells us something about deterministic computations that $P \neq NP$ doesn't already say; intuitively, this is hard to grasp.