Consider a first-order theory having a unary function symbol $f$. Then the following quasi-identity expresses that $f$ is injective.
$$\forall xy : f(x)=f(y) \rightarrow x=y$$
Alternatively, we can demand that $f$ be surjective by adjoining the following axiom.
$$\forall y\exists x : f(x)=y$$
Observe, however, that this isn't a quasi-identity.
Question 0. Is there a quasi-identity expressing that $f$ is surjective?
Question 1. If not, are sentences of the form [Universal quantifiers][Existential quantifiers][Equation] somehow "dual" to quasi-identities?
No, there is no quasi-identity that will do it.
Let $\mathcal{C}$ be the category of structures for your signature, i.e. pairs of sets equipped with a map between them (or if you prefer, sets equipped with an endomap; the same argument works for both). Let $\mathcal{A}$ be the full subcategory of structures satisfying some given quasi-identities. It is not hard to see that $\mathcal{A}$ is closed under (possibly infinite) products and subobjects. But this is not true for the subcategory where the equipped map is surjective. (Infinite products are a non-issue if you have the axiom of choice – but there are easy counterexamples to the closure under subobjects.)