YES

We show the termination of R/S, where R is

  f(b(x),y) -> f(x,b(y))
  f(x,a(y)) -> f(a(x),y)

and S is:

  f(x,y) -> f(a(x),y)
  f(x,y) -> f(x,b(y))

Since R almost dominates S and S is non-duplicating,
it is enough to show finiteness of (P, Q).
Here P consists of the dependency pairs

  f#(b(x),y) -> f#(x,b(y))
  f#(x,a(y)) -> f#(a(x),y)

and Q consists of the rules:

  f(b(x),y) -> f(x,b(y))
  f(x,a(y)) -> f(a(x),y)
  f(x,y) -> f(a(x),y)
  f(x,y) -> f(x,b(y))
  f#(x,y) -> f#(a(x),y)
  f#(x,y) -> f#(x,b(y))

The weakly monotone algebra (N^3, >_lex) with

  f#_A((x1,y1,z1),(x2,y2,z2)) = (x1, 0, 0)
  b_A((x1,y1,z1)) = (x1 + 1, y1 + 1, z1 + 1)
  a_A((x1,y1,z1)) = (0, 1, 1)
  f_A((x1,y1,z1),(x2,y2,z2)) = (0, 0, 0)

strictly orients the following dependency pairs:

  f#(b(x),y) -> f#(x,b(y))

We remove them from the set of dependency pairs.

The weakly monotone algebra (N^3, >_lex) with

  f#_A((x1,y1,z1),(x2,y2,z2)) = (x2, y2, z2)
  a_A((x1,y1,z1)) = (x1, y1 + 1, z1 + 1)
  f_A((x1,y1,z1),(x2,y2,z2)) = (0, 0, 0)
  b_A((x1,y1,z1)) = (0, 0, 0)

strictly orients the following dependency pairs:

  f#(x,a(y)) -> f#(a(x),y)

We remove them from the set of dependency pairs.

No dependency pair remains.