What is so special about $a*b^{ -1}$ equivalence?

I presume you mean the equivalence relation induced by a subgroup $H\le G$ defined by $a\sim b$ if and only if $ab^{-1}\in H$. I have two answers for this.

(1) Groups are meant to study symmetry. I think groups are to group actions as potential energy is to kinetic energy: groups act on things. So group actions are of fundamental importance. Often we examine certain types of actions on certain types of objects: linear transformations on a vector space, or automorphisms of a graph, or rotations preserving a polyhedron, or diffeomorphisms of a manifold, and so on. But the most basic is just a group action on a set, with no additional structure involved. This type of thing is used in leveraging symmetry to explore combinatorial counting problems, for example necklace counting. This is relevant to Polya enumeration, cycle indices, generating functions, and most generally combinatorial species.

Anyway, the most basic building block in the theory of group actions is that of orbits. We can say that every $G$-set is a disjoint union of orbits. The categorical version of the orbit-stabilizer theorem states that every orbit is isomorphic (as a $G$-set, via a $G$-equivariant bijection) to the coset space $G/H$, where $H$ is the stabilizer of any element in the orbit. So coset spaces are natural, and they form a partition of $G$. Partitions of sets correspond to equivalence relations, and the equivalence relation in this case is $aH=bH$, or equivalently $b^{-1}a\in H$. With right actions it's $ab^{-1}\in H$.

(2) Congruence relations are fundamental in abstract algebra for collapsing algebraic structures into quotients (quotient groups, quotient rings, quotient modules, etc.), and by the first isomorphism theorem in the appropriate category this is equivalent to characterizing all of the homomorphic images of an algebraic structure. A congruence relation $\sim$ on a group $G$ is an equivalence relation such that $a\sim b$ and $c\sim d$ imply $ac\sim bd$ for all $a,b,c,d\in G$. One may prove that such a relation is precisely that of $a\sim b$ iff $aN=bN$ for some normal subgroup $N\triangleleft G$.

I presume when you say "equivalence $a \cdot b^{-1}$" you mean a relation $a \cdot b^{-1} = 1$ in a group presentation. (We may also consider a more general situation in which we impose the relation $a \cdot b^{-1}$.) It is not clear from your question wheter $a$ and $b$ should be generators or arbitrary expressions, but that doesn't matter too much.

In group theory can we present finitely generated congruences by elements because we can always transform an equation $L = R$ to one of the form $L \cdot R^{-1} = 1$. If you meant for $a$ and $b$ to be arbitrary expressions, then the answer is: because group theorists "optimize" group presentations so that they are of the form $E = 1$ where $E$ is an expression. They could not afford to do it if they worked with semigroups, for instance. This kind of "optimization" can be harmful to students if they are not told how things fit into a more general picture: a group presentation is just a quotient of a free group by a congruence relation. There are billions of ways of defining a congruence relation, of which group theorists particularly like finite presentations, and they use the group structure to always present the relations in the form $E_1 = 1, \ldots, E_n = 1$ – and they just list $E_1, \ldots, E_n$.

If $a$ and $b$ are supposed to be generators, or particular elements of a group (such as in the case of getting $\mathbb{Z}/n$ from $\mathbb{Z}$), then the answer is again similar: the congruence relation generated by $a \cdot b^{-1}$ is just imposing the equation $a = b$, which just means "let us make $a$ and $b$ equal". Obviously, there will be many applications in which we would want to equate two elements. Again, it is an accident of group theory that equating $a$ and $b$ is the same as equating $a \cdot b^{-1}$ and $1$.

The whole thing reminds me a bit of how students of homological algebra are not aware of the connection between exact sequences and coequalizers (because nobody told them): in an abelian category the coequalizer $q : B \to Q$ of $f : A \to B$ and $g : A \to B$ can be "optimized" to an exact sequence $$A \to B \to Q \to 0 $$ where the arrow $A \to B$ is $f - g$ and the arrow $B \to Q$ is $q$. Again, we have the same idea: rather than equating $f(x) = g(x)$ we equate $f(x) - g(x) = 0$. And the students are only ever shown the exact sequence, which is a shame.