YES

Problem:
 +(0(),y) -> y
 +(x,0()) -> x
 +(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:
  +(0(),y) -> y
  +(x,0()) -> x
  +(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:+(0(),y) -> y
         +(x,0()) -> x
         +(s(x),y) -> s(+(x,y))
         +(x,s(y)) -> s(+(x,y))
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  0() = 0(), s(y) = s(+(0(),y)), s(x) = s(+(x,0())), s(+(x186,0())) = 
  s(x186), s(+(x188,s(y))) = s(+(s(x188),y)), s(+(0(),x191)) = s(x191), 
  s(+(s(x),x193)) = s(+(x,s(x193)))
  
   CP(S,P union P^-1) = 
  y = +(y,0()), x = +(x,0()), x = +(0(),x), y = +(0(),y), s(+(x210,y)) = 
  +(y,s(x210)), s(+(x212,x)) = +(x,s(x212)), s(+(x,x215)) = +(s(x215),x), 
  s(+(y,x217)) = +(s(x217),y)
  
   CP(P union P^-1,S) = 
  +(y,0()) = y, +(0(),x) = x, +(y,s(x)) = s(+(x,y)), +(s(y),x) = s(+(x,y))
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [+](x0, x1) = x0 + x1 + 6,
    
    [0] = 1,
    
    [s](x0) = x0 + 1
   orientation:
    +(0(),y) = y + 7 >= y = y
    
    +(x,0()) = x + 7 >= x = x
    
    +(s(x),y) = x + y + 7 >= x + y + 7 = s(+(x,y))
    
    +(x,s(y)) = x + y + 7 >= x + y + 7 = s(+(x,y))
   problem:
    +(s(x),y) -> s(+(x,y))
    +(x,s(y)) -> s(+(x,y))
   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