Reducing the Hamiltonian cycle to the travelling salesman problem and self loops
Suppose your original graph $G$ has a Hamiltonian cycle $C$.
Then the cost of the tour induced by $C$ in the new graph $G'$ you defined is indeed $0$ and conversely, any TSP tour of cost $0$ can only use edges of cost $0$ as you did not introduce any edges with a negative cost.
Adding the loops does not change this, so the reduction works out if we are willing to allow non-simple graphs which isn't done usually as the definition of the TSP asks for a tour visiting each vertex exactly once, i.e. no valid tour would include loop edges anyways.
Given this, we also see that you can remove the loop edges in $G'$ to make it a simple graph to obtain an equally valid reduction.
From a technical standpoint this doesn't make much of a difference but since people like simple graphs, I would personally prefer the reduction yielding a simple graph.