 Solution

Use $c=[a,b]=a^{-1}{b}^{-1}ab$. The two defining relators say that $c$ commutes with $a,b$, hence is central. The given group commutator identities giveConsequently a word collection in a two-step nilpotent group algorithm maintains an integer triple $(m,n,k)$, initially $(0,0,0)$. Reading $a^{ϵ}$, with $ϵ=\pm 1$, changes it to $(m+ϵ,n,k-nϵ)$; reading $b^{ϵ}$ changes it to $(m,n+ϵ,k)$. Each step uses the displayed multiplication rule and reads one letter, so it terminates and produces the requested expression.

By the stipulated uniqueness, the word is trivial exactly when $\boxed{(m,n,k)=(0,0,0)}$, which solves its word problem for a group. Moreover the commutator subgroup is $⟨c⟩$: the quotient by $⟨c⟩$ is abelian, and $c$ itself is a commutator. It is central, so the nilpotency class is at most two. Uniqueness also implies $c\ne 1$, since $(0,0,1)$ and $(0,0,0)$ are distinct triples. Hence the exact nilpotency class is $\boxed{2}$.