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, >) with

  f#_A(x1,x2) = x1
  b_A(x1) = x1 + 1
  a_A(x1) = x1
  f_A(x1,x2) = x1

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, >) with

  f#_A(x1,x2) = x2
  a_A(x1) = x1 + 1
  f_A(x1,x2) = x2
  b_A(x1) = x1

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.