There's a relation $\rho$ defined as $\forall a,b \in \mathcal{A}$ such that $\mathcal{A} = R - \{0\} $ and $(x,y)\in \rho = x \cdot y > 0$. I proved that this is in fact an equivalent relation and I went up to construct all the equivalent sets.
I came up with the first one, that is:
$\bar{1} = \{x \in \mathcal{A} \text{ | } (x,1)\in \mathcal{A} \} = \{x \in \mathcal{A} \text{ | } x > 0\}$
From what I did, I would say that I have half of all the numbers in the original set $\mathcal{A}$. I tried to construct the class of $\overline{-1}$ but I currently don't know if it is possible.
The classes must partition $\mathbb{R}-\{0\}$, so make use of that. You've correctly identified $\overline{1}$ as $(0, +\infty)$. That isn't all of $\mathbb{R}-\{0\}$, so there must be another class. Take an element not in $\overline{1}$, maybe $-1$. Then $x \sim -1$ if and only if $-x >0$, which is true if and only if $x < 0$. Hence $\overline{-1}=(-\infty, 0)$. Since these two classes cover all of $\mathbb{R}-\{0\}$, you're done.