Let's look at an example. Suppose you have four cities, A, X, Y, and Z, and A is at distance $1$ from each of the others, while the distance between any two of X, Y, Z is $100$. Then of course a solution to TSP starting at A is given by A to X to A to Y to A to Z to A – let's write this as AXAYAZA, which has length $6$, and visits A multiple times.
Now let's carry out the construction in Rahul's comment. We form a new problem with the same cities but with the distance between cities replaced by the length of the shortest path between the two cities. So in the new problem, A is still at distance $1$ from each of the others, but the distance between any two of X, Y, Z is $2$. Now the tirangle inequality is satisfied (although the configuration is still unrealizable in any Euclidean space), and a solution is given by AXYZA, length $6$, no city visited twice (except for both starting and ending at A).
So we have taken a problem where the solution has multiple visits to a city, and replaced it with an equivalent problem which doesn't. I hope it's clear that you can carry out this construction for any (finite) set of cities and distances. So you can always replace a problem with an equivalent problem where the triangle inequality is satisfied and no city gets visited more than once. So, there is no loss of generality in assuming the triangle inequality holds and no city gets visited more than once. That's why we can restrict our attention to what looks like, but really isn't, a special case.
This variant is known as the Set TSP, as well as several other names mentioned in the link.
The first reference (Noon and Bean, 1993) describes a transformation in which each set/type is visited exactly once but mentions that Noon's dissertation [10] shows how to handle intraset arcs.
Best Answer
I'm not sure if there is an exact name for this, but this problem is certainly NP hard. We see that the absolute shortest possible way to do this is to visit each vertex exactly once, which is precisely the Hamiltonian Path problem, which is known to be NP complete. Hence, your problem is going to be in NP, as if the shortest is everything exactly once, we have solved Hamiltonian Path.