You will get (C,D,E,A,B) or (C,D,E,B,A).

I added an example of DFS pseudo code below.

When you check line 5 ( the line with underline), you could see a for loop that iterates all vertex in the graph. 

So,

(C,D,E,A,B)   // if the for loop iterated vertex A after we went through C,D,E

(C,D,E,B,A)  //  if the for loop iterated vertex B after we went through C,D,E 

---------------------------------------------------

 DFS (G)
1 for each vertex u in G.V
2     u:color = WHITE
3     u:.parent = NIL
4 time = 0
5 for each vertex u in G.V
6     if u.color == WHITE
7         DFS-VISIT(G,u)

DFS-VISIT(G, u)
1 time = time + 1             // white vertex u has just been discovered
2 u.d = time
3 u.color = GRAY
4 for each v in G.Adj[u]    // explore edge (u,v)
5     if v.color == WHITE
6         v.parent = u
7         DFS-VISIT(G,v)
8 u.color = BLACK            // blacken u; it is finished
9 time = time + 1
10 u.f = time

-------------------------------------

Thanks for clarifying.

I ran into this question because on the code provided on wiki, if I try this graph with edges I provide, I get (C, D, E).

A DFS will get you (C, D, E)

A DFS-loop will get you (C, D, E), then (A, B)

Or (C, D, E), then (B), and finally (A)

It really depends which node the loop gets to after it's done exploring everything it can starting from C

The important thing to take away from this is that a DFS (and a BFS for that matter) will get to everything reachable from the start node and nothing else

This is the point of the first-pass DFS; we need the proper search order of the nodes so a search will only visit 'sink' SCCs.   Why it accomplishes this, you'll have to watch and understand the proof of Karasuja's algorithm.