The proof of Lemma 4 is incorrectly formulated. It shall be as follows.

We define B as in the paper. f is defined by induction on the length of a word
in the following way: f(x)=x for every x\in A, f(xwx^{-1})=x.f(w) provided that
w does not start with x^{-1} and f(w) is defined, f(x^{-1})=x\f(w) provided that
w does not start with x and f(w) is defined.

We need to prove that f(u*v)=f(u).f(v) for every u,v\in B. We proceed by
induction on length of u. 

(I) If |u|=1, there are two cases. Either v starts
with u. In this case, either v=u^{-1}wu in reduced form
and so f(u*v)=f(w), while f(u).f(v)=u.f(v)=u(u\f(w))=f(w)
by definition of f, or v=u and then the claim follows from idempotency of A. 
Or v does not start with v and we can use the definition of f.
Similarly, one can prove the following claim (I'): For every x\in A, v\in B, 
f(x^{-1}*v)=x\f(v).

(II) Assume that f(u*v)=f(u).f(v) for every u,v with |u|<=n. We prove that f(u*v)=f(u).f(v) 
for every u,v with |u|=n+2. Such u is uniquely expressed as xwx^{-1} or x^{-1}wx, where 
x\in A, |u|=n. We will handle the first case, the second is similar. Then f(u*v)=
f(x*(w*(x^{-1}*v)))=x.f(w*(x^{-1}*v)) by (I), =x.(f(w).f(x^{-1}*v)) by induction assumption,
=x.(f(w).(x\f(v))) by (I'), =(x.f(w)) . (x.(x\f(v)) by left distributivity and this is
f(u).f(v) by definition of f and cancellation.

****

Later, I found that Theorem 6 is not new. It was proved in
S. Fajtlowicz and J. Mycielski, On convex linear forms. Algebra Universalis 4 (1974), 244--249.