If you allow equality then it is trivial that every graph has a distinguishing famil of properties. Since you are considering the property "$v$ has exactly one neighbor", I think you are including equality...
Indeed, let $G$ be a graph, let $v_1,\dots,v_n$ be its vertices, and let me construct a 'distinguishing property' $\phi$ for $v_1$: put $\phi=(\exists x_2,x_2,\dots,x_n)\Phi(v_1,x_2,\dots,x_n)$ with $\Phi(x_1,\dots,x_n)$ being the formula which says that all its arguments are distinct and that the $i$th and $j$th arguments are $\sim$-related iff the vertices $v_i$ and $v_j$ are connected in the graph $G$.
Using formulas constructed like this, I think you can answer your (main?) question.