Is Scott's reduction sound for $\mathrm{FO}^2$ with equality?
Yes. Just employ the formula $\forall{x}\forall{y} \; R(x,y) \leftrightarrow (x=y)$ (for a fresh binary predicate $R$), which allows you to &quot;hide&quot; the equality inside the $\forall\forall$-part of the Scott normal form. Then you proceed as usual.
EDIT: I've noticed that you wrote that $\alpha$ in $\forall{x}\forall{y} \; \alpha$ is a binary predicate. It is not true and obviously not all $\textit{FO}^2$ formulae are reducible to such a form. The aforementioned $\alpha$ should be a quantifier-free formula (line 13 of page 60 of the attached pdf). Maybe this is what made you problems while understanding the Scott reduction.