You are right – you cannot simply return the stack, it indeed contains a lot of unvisited nodes.
However, by maintaining a map (dictionary): map:Vertex->Vertex such that parentMap[v] = the vertex we used to discover v, you can get your path.
The modification you will need to do is pretty much in the for loop:
for child in children:
stack.push(child[0])
parentMap[child] = parent #this line was added
Later on, when you found your target, you can get the path from the source to the target (pseudo code):
curr = target
while (curr != None):
print curr
curr = parentMap[curr]
Note that the order will be reversed, it can be solved by pushing all elements to a stack and then print.
I once answered a similar (though not identical IMO) question regarding finding the actual path in BFS in this thread
Another solution is to use a recursive version of DFS rather then iterative+stack, and once a target is found, print all current nodes in the recursion back up – but this solution requires a redesign of the algorithm to a recursive one.
P.S. Note that DFS might fail to find a path to the target (even if maintaining a visited set) if the graph contains an infinite branch.
If you want a complete (always finds a solution if one exists) and optimal (finds shortest path) algorithm – you might want to use BFS or Iterative Deepening DFS or even A* Algorithm if you have some heuristic function