Of course if $\pi_0$ is defined by the equivalence relation generated by ~ on $0$-simplices then it is the usual thing: topological $pi_0$ \pi_0$ of the realization, or simplicial $pi_0$ \pi_0$ of a fibrant replacement.
You are sayingwhat : What if we define a new $\pi_n(A,x)$ as $\pi_0(A(n,x))$? Well, obviously it maps to the usual $\pi_n(A,x)=\pi_0(S(A),n,x)$), and clearly this map is rarely an isomorphism if $A$ is not fibrant. But I don't even see a comparison map between the resulting homotopy category and the usual one, in either direction. Clearly a map of simplicial sets will sometimes induce an isomorphism of usual homotopy groups while inducing a non-isomorphism of yours. But (this is my point) it can also go the other way. For example, you can make lots of examples of simplicial sets $A$ such that the inclusion $V\to A$ of the $0$-skeleton of $A$ induces an isomorphism $V(n,x)\to A(n,x)$.