$\mathrm{Sym}(\{0,1,2\})\cong ⟨s,t\mid s^2,t^2,(st{)}^3⟩$

Direct permutation computation gives $s^2=t^2=\mathrm{id}$ and $(st{)}^3=\mathrm{id}$, so [L2] constructs a homomorphism $\pi :P\to \mathrm{Sym}(A)$.

After cancelling $s^2$ and $t^2$, every word alternates. The relation $(st{)}^3=e$ gives $stst=ts$ and hence $tst=sts$ after multiplying on the left by $s$; replacing the first three letters of any alternating word of length at least four by the other side creates an adjacent equal pair and shortens the word.

Their images are respectively $\mathrm{id},(01),(12),(012),(021),(02)$, so they are distinct; [L1] gives $\mathrm{\mid }\mathrm{Sym}(A)\mathrm{\mid }=3!=6$, and these images exhaust it.

Repeating step 1.2 leaves one of $e,s,t,st,ts,sts$, since the two alternating words of length three are equal.

Step 2.1 gives at most six elements in $P$, while step 1.3 gives six distinct images under $\pi$; hence $\pi$ is bijective and is the claimed isomorphism.