Recursive string substitution

Python 2, 43 bytes

lambda s,*l:eval('s'+'.replace(*l)'*len(s))

Try it online!

Evaluates a string of the form

s.replace(*l).replace(*l).replace(*l) ...

To reach a fixed point if one exists, it suffices to do replacements equal to the length of the original string.

05AB1E, 2 bytes

`:

Try it online!

Explanation

`    # split input to stack
 :   # replace (until string doesn't change)

This could be : for 1 byte if we didn't have to deal with empty strings.

ES6 (Javascript), 47, 43 bytes

• Saved 4 bytes using currying (Thanks @Neil !)

Golfed

c=>b=>R=a=>(x=a.split(b).join(c))==a?x:R(x)

Try It

Q=c=>b=>R=a=>(x=a.split(b).join(c))==a?x:R(x)

function doit() {
  console.log(Q(C.value)(B.value)(A.value));
}

A: <input type="text" value="abaaba" id="A"/> B: <input type="text" value="aba" id="B"/> C: <input type="text" value="ab" id="C"/> <input type="submit" onclick="doit();" value="REPLACE"/>