 Solution

For the additive zero singleton, $x+x=x$ is already a quantifier-free formula, so it is also an existential formula and a universal formula with an empty quantifier block. If a nonempty block is required, add a dummy $\exists y$ or $\forall y$; first-order domains are nonempty.

For the two definable multiplicative singletons, existential formulas areIn the integers, the equation $x^2=x$ restricts $x$ to zero or one. In the first line, one never satisfies the inequality, whereas zero does with $y=1$. In the second, zero never satisfies it, whereas one does with $y=0$. The respective universal formulas are $\forall y(xy=x)$ and $\forall y(xy=y)$, as already verified.

Neither multiplicative singleton has a quantifier-free formula. In this constant-free, multiplication-only language, every term in the sole free variable $x$ is $x^k$ for a positive integer $k$. Every atomic formula $x^r=x^s$ is true at both zero and one. Every Boolean combination consequently has the same truth value at those two points, and cannot separate their singletons. This is the fact that zero and one have the same quantifier-free type in the multiplicative integers.

Thus additive zero admits all three forms; multiplicative zero and one admit existential and universal forms, but no quantifier-free form. All remaining cases were excluded from definability itself by structure automorphisms.