YES

Problem:
 +(s(x),y) -> s(+(x,y))
 +(x,s(y)) -> s(+(x,y))
 +(x,y) -> +(y,x)

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  +(s(x),y) -> s(+(x,y))
  +(x,s(y)) -> s(+(x,y))
  +(x,y) -> +(y,x)
  
   T' = (P union S) with
  
   TRS P:+(x,y) -> +(y,x)
  
   TRS S:+(s(x),y) -> s(+(x,y))
         +(x,s(y)) -> s(+(x,y))
  
  S is left-linear and P is reversible.
  
   CP(S,S) = 
  s(+(x62,s(y))) = s(+(s(x62),y)), s(+(s(x),x65)) = s(+(x,s(x65)))
  
   CP(S,P union P^-1) = 
  s(+(x74,y)) = +(y,s(x74)), s(+(x76,x)) = +(x,s(x76)), s(+(x,x79)) = 
  +(s(x79),x), s(+(y,x81)) = +(s(x81),y)
  
  
   PCP_in(P union P^-1,S) = 
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [+](x0, x1) = x0 + 2x1 + 7,
    
    [s](x0) = x0 + 6
   orientation:
    +(s(x),y) = x + 2y + 13 >= x + 2y + 13 = s(+(x,y))
    
    +(x,s(y)) = x + 2y + 19 >= x + 2y + 13 = s(+(x,y))
   problem:
    +(s(x),y) -> s(+(x,y))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [+](x0, x1) = 6x0 + x1 + 2,
     
     [s](x0) = x0 + 3
    orientation:
     +(s(x),y) = 6x + y + 20 >= 6x + y + 5 = s(+(x,y))
    problem:
     
    Qed