Is there an algorithm that can solve any two-player game?
(Note that you're excluding games with randomness, such as backgammon.)

Yes. In particular, endgame tablebases use most of such an algorithm.

Wbool,Wfull,Bbool,Bfull = set(),set(),set(),set()
keep_going = False
for v in nondraw_endstates:
 keep_going = True
 if is_a_victory_for_white_(v):
  Wbool.insert(v)
  Wfull.insert([v,0])
 else:
  Bbool.insert(v)
  Bfull.insert([v,0])
n = 1
while keep_going:
 keep_going = False
 for v in Wbool:
  for u in nodes_with_an_edge_to_(v):
   if u in Wbool:
    continue
   if white's_turn_(u):
    Wbool.insert(u)
    Wfull.insert([u,n,v])
    keep_going = True
    continue
   avoidable = False
   for x in nodes_with_an_edge_from_(u):
    if x not in Wbool:
     avoidable = True
     break
   if not avoidable:
    Wbool.insert(u)
    Wfull.insert([u,n,v])
    keep_going = True
 for v in Bbool:
  for u in nodes_with_an_edge_to_(v):
   if u in Bbool:
    continue
   if not white's_turn(u):
    Bbool.insert(u)
    Bfull.insert([u,n,v])
    keep_going = True
    continue
   avoidable = False
   for x in nodes_with_an_edge_from_(u):
    if x not in Bbool:
     avoidable = True
     break
    if not avoidable:
     Bbool.insert(u)
     Bfull.insert([u,n,v])
     keep_going = True
     continue
 if (initial_node in Wbool) or (inital_node in Bbool):
  keep_going = False
 else:
  n = n+1


Suppose the game is locally finite. If white has a winning strategy then the while loop will terminate and every extension of the result of removing the middle entries from Wfull's triples to all non-endstate white nodes is a winning strategy for white. ​ If black has a winning strategy then the while loop will terminate and every extension of the result of removing the middle entries from Bfull's triples to all non-endstate black nodes is a winning strategy for black. If only finitely many nodes can reach the non-draw endstates and neither side has a winning strategy then the while loop will terminate and white can always stay out of Bbool and white doing so guarantees that white won't lose. Bblack can always stay out of Wbool and black doing so guarantees that black won't lose.

If the while loop terminates, then it does so in an amount of time that's bounded above by a polynomial in (the number of nodes that can reach the non-draw endstates). Furthermore, by reduction from the circuit value problem, deciding who has a winning strategy is P-complete even when $G$ is (finite and) acyclic and there are no draw states.