= 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> are
$$
\boxed{\begin{aligned}
0:\quad&\exists y\,(x\cdot x=x\ \land\ x\cdot y\ne y),\\
1:\quad&\exists y\,(x\cdot x=x\ \land\ x\cdot y\ne x).
\end{aligned}}
$$
In 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 \b[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>.
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