Rewrite the given condition successively $(x + y)^3 - 3xy(x + y) = x^3y^3 - 3x^2y^2$, i.e. $(x + y)^3 - (xy)^3 = 3xy(x + y) - 3x^2y^2$, or $(x + y - xy)(x^2 + 2xy + y^2 + x^2y + xy^2 + x^2y^2) = 3xy(x + y - xy)$. We either have $x + y = xy$, which leads to $E = 1$ (obtained for $x = y = 2$), or $x^2 + 2xy + y^2 + x^2y + xy^2 + x^2y^2 = 3xy$. The last equality, multiplied by $2$, can be written equivalently $x^2(y + 1)^2 + y^2(x + 1)^2 + (x - y)^2 = 0$, which is possible if and only if $x = y = -1$ ($x = y = 0$ is forbidden). In this case, $E = -2$. In conclusion, the set of the possible values of $E$ is $\{-2, 1\}$.