YES

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

  f(el(x),y) -> f(x,el(y))

and S is:

  f(x,y) -> f(l(x),y)
  f(x,y) -> f(x,r(y))
  l(el(x)) -> el(l(x))
  el(r(x)) -> r(el(x))

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

and Q consists of the rules:

  f(el(x),y) -> f(x,el(y))
  f(x,y) -> f(l(x),y)
  f(x,y) -> f(x,r(y))
  l(el(x)) -> el(l(x))
  el(r(x)) -> r(el(x))
  f#(x,y) -> f#(l(x),y)
  f#(x,y) -> f#(x,r(y))

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

  f#_A((x1,y1,z1),(x2,y2,z2)) = (x1, 0, 0)
  el_A((x1,y1,z1)) = (x1 + 1, 1, 1)
  f_A((x1,y1,z1),(x2,y2,z2)) = (x1, 0, 0)
  l_A((x1,y1,z1)) = (x1, y1 + 1, 1)
  r_A((x1,y1,z1)) = (1, 2, 2)

strictly orients the following dependency pairs:

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

We remove them from the set of dependency pairs.

No dependency pair remains.