I know that finding the optimal solution to One Way TSP (TSP but the salesman does not have to return to his original city) is NP-Hard, but is it NP-Complete? I ask this because I recently found a solution to Open TSP but can't find a good resource to tell me whether or not One Way TSP is NP-Complete.
The answer depends on how you define "One Way TSP".
If the One Way TSP problem asks to compute the tour itself, then it cannot possibly be NP-complete since it is not a decision problem, and hence it does not even belong to the class NP.
If the problem is that of deciding whether there is a tour having cost at most $x$, for some input parameter $x$, then the problem belongs to NP (a yes-certificate is the tour itself) and it is NP-complete.
Take your "one way" problem. Add another city X, make the distance from start to X = 0, make the distance from X to any other city d which is longer than the sum of all other distances, then solve the original TSP for this instance.
In the solution, X obviously must have two neighbours. One neighbour must be the start city, otherwise you would have a total distance >= 2d, instead of a distance less than 2d. If you remove the city X from the solution, you have the optimal solution for the "one way" problem.