A balanced tree-like presentation of $S_3$ Does the 6-element group $S_3$ have a finite (balanced) semigroup presentation of the form $$\langle a_1,...,a_n\mid  a_1=u_1, a_2=u_2,...,a_n=u_n\rangle$$ where $u_1,u_2,...,u_n$ are semigroup words? Let us call such a  semigroup presentation {\it tree-like}.
Edit: By a semigroup word, I mean a word without inverses of letters. By a semigroup presentation, I mean a presentation in the class of semigroups (not groups). For example, $\langle a,b\mid a^3=a, b=ab^2a\rangle$ defines $S_3$ in the class of groups but a "bigger" semigroup in the class of semigroups. In that semigroup, $b$ does not divide $a$, so it is not a group.
Motivation 1. The cyclic group of order $n$ has a tree-like semigroup presentation $\langle a\mid a=a^{n+1}\rangle$. Matt Brin noted that the 8-element quaternion group has the tree-like presentation $\langle a,b,c \mid a=bc, b=ca, c=ab\rangle$. The groups $S_m, m>3$ do not have tree-like presentations because they do not even have balanced (same number of relations and generators) presentations at all since their Schur multipliers are non-trivial. The group $S_3$ has a balanced presentation.
Motivation 2. Every tree-like semigroup presentation corresponds to a (closed) subgroup of the R. Thompson group $F$.  For example as shown by Guba the Brin tree-like presentation of the quaternion group corresponds to a copy of the Thompson group $F_9$ (the group of piecewise linear homeomorphisms of $[0,1]$ with slopes of the form $9^k$ and break points of the derivative from $\mathbb{Z}[1/9]$).
 A: Campbell, C. M.; Mitchell, J. D.; Ruškuc, N. On defining groups efficiently without using inverses. Math. Proc. Cambridge Philos. Soc. 133 (2002), no. 1, 31–36 shows among other things that a group with a balanced group presentation has a balanced semigroup presentation. The proof gives a tree like presentation. Look at the proof of Prop 2.4 and 2.5 and note R' and R'' are empty when the group presentation was balanced.  See https://www.cambridge.org/core/journals/mathematical-proceedings-of-the-cambridge-philosophical-society/article/on-defining-groups-efficiently-without-using-inverses/ED8A580F336B47B2162DB5E3A5FA8459
A: I see that Jeremy has beaten me to it - but here is a solution found by computer by choosing random entries from the group multiplication table.
This one also generates $S_3$ as a semigroup. The Magma command $\mathtt{RWSMonoid}$ applies the Knuth-Bendix algorithm to the presentation, and regards it as a monoid presentation. It has order $7$ as a monoid with identity equal to the empty word, and so it has order $6$ as a semigroup presentation.
$$\langle a,b,c,d,e,f\mid a=f^2,b=ab,c=df,d=ec,e=db,f=ce\rangle.$$
  > G := Group< a,b,c,d,e,f | a = f^2, b = a*b, c = d*f, d=e*c, e=d*b, f=c*e >;
  > Order(G);
  6
  > Homomorphisms(G,Sym(3))[1];
   Homomorphism of GrpFP: G into GrpPerm: X, Degree 3, Order 2 * 3 induced by
      G.1 |--> Id(X)
      G.2 |--> (1, 2, 3)
      G.3 |--> (1, 3, 2)
      G.4 |--> (1, 2)
      G.5 |--> (1, 3)
      G.6 |--> (2, 3)

> M := Monoid< a,b,c,d,e,f | a = f^2, b = a*b, c = d*f, d=e*c, e=d*b, f=c*e >; 
> R := RWSMonoid(M);
> Order(R);
7

