Point in polygon algorithm that handles polygons that straddle the antemeridian

Standard convex hull algorithms are not defeated by the wrapping-around of the coordinates on the surface of the Earth but by a more fundamental problem. The surface of a sphere (let's forget the not-quite-sphericity of the Earth) is not a Euclidean space so Euclidean geometry doesn't work, and convex hull routines which assume that the underlying space is Euclidean (show me one which doesn't, please) won't work.

The surface of the sphere conforms to the concepts of an elliptic geometry where lines are great circles and antipodal points are considered the same point. You've already started to experience the issues arising from trying to apply a Euclidean concept of convexity to an elliptic space.

One approach open to you would be to adopt the definitions of geodesic convexity and implement a geodesic convex hull routine. That looks quite hairy. And it may not produce results which conform to your (generally Euclidean) expectations. In many cases, for 3 arbitrary points, the convex hull turns out to be the entire surface of the sphere.

Another approach, one adopted by navigators and cartographers through the ages, would be to project part of the surface of the sphere (a part containing all your points) into Euclidean space (which is the subject of map projections and I won't bother you with references to the extensive literature thereon) and to figure out the convex hull of the projected points. Project the area you are interested in onto the plane and adjust the coordinates so that they do not wrap around; for example, if you were interested in France you might adjust all longitudes by adding 30deg so that the whole country was coordinated by +ve numbers.

While I'm writing, the idea proposed in @Li-aung Yip's answer, of using a 3D convex hull algorithm, strikes me as misguided. The 3D convex hull of the set of surface points will include points, edges and faces which lie inside the sphere. These literally do not exist on the 2D surface of the sphere and only change your difficulties from wrestling with the not-quite-right concept in 2D to quite-wrong in 3D. Further, I learned from the Wikipedia article I referenced that a closed hemisphere (ie one which includes its 'equator') is not convex in the geometry of the surface of the sphere.