The answer to your question is a qualified no. Part of the reason is that we can't assume that every object in our model is named by an individual constant. So for instance, it could be that our model satisfies the sentence $\bigwedge_{i \in I} \neg P(c_i)$, where "$\bigwedge_{i \in I}$" indicates (possibly infinite) conjunction over index set $I$, and $c_i$ are all constants of the language, and yet this same model also satisfies the sentence $\exists x P(x)$. It's just that the object in our model which satisfies $P(x)$ is unnamed.
Of course, you're right that there is a strong analogy between quantifiers and infinite conjunctions/disjunctions in the following sense: if we require that every object in our domain is named by a constant, and if we allow for arbitrary conjuncts/disjuncts, then we can translate the quantified sentences into quantifier-free sentences using (possibly infinite) conjunctions/disjunctions. Logicians sometimes define substitutional quantifiers for this purpose: for instance, letting $\Sigma$ be a new substitutional quantifier, $\Sigma x \varphi(x)$ is true in a model just in case for some constant $c$, $\varphi(c)$ is true in that model, i.e. just in case $\bigvee_{i \in I} \varphi(c_i)$ is true in that model, where $I$ indexes the constants of $L$.
With that said, an infinitary propositional logic without quantifiers is not the same as a first-order logic with quantifiers. For one thing, in a propositional logic, you can only say $p$ is true or false. Your models aren't collections of objects with structure, but rather are simply truth value assignments for the proposition letters. So it's hard to say in what sense, if any, an infinitary propositional logic is the same as first-order logic without infinitary conjunctions/disjunctions. Their models don't even look alike.
Furthermore, even an infinitary predicate logic without quantifiers fails to be equivalent to first-order logic with quantifiers (but only finite conjunctions/disjunctions). The reason is simple: in first-order logic, there is no sentence which is true exactly when the domain is infinite. However, if the language you invoke has (at least) countably many constants $c_i$, then the sentence $\bigwedge_{\substack{i,j \in \omega \\ i \neq j}} c_i \neq c_j$ can only be true in infinite models. Hence, infinitary predicate logic without quantifiers is not compact, and so can't be equivalent to first-order logic.
here are some basic distributive properties in quantifiers, hope it might help someone.
∀x(P(x) ∧ Q(x)) ≡ (∀xP(x) ∧ ∀xQ(x))
∃x(P(x) ∧ Q(x)) → (∃xP(x) ∧ ∃xQ(x))
∀x(P(x) ∨ Q(x)) ← (∀xP(x) ∨ ∀xQ(x))
∃x(P(x) ∨ Q(x)) ≡ (∃xP(x) ∨ ∃xQ(x))
∀x(P(x) → Q(x)) ← (∃xP(x) → ∀xQ(x))
∃x(P(x) → Q(x)) ≡ (∀xP(x) → ∃xQ(x))
∀x¬P(x) ≡ ¬∃xP(x)
∃x¬P(x) ≡ ¬∀xP(x)
∀x∃yT(x,y) ← ∃y∀xT(x,y)
∀x∀yT(x,y) ≡ ∀y∀xT(x,y)
∃x∃yT(x,y) ≡ ∃y∃xT(x,y)
∀x(P(x) ∨ R) ≡ (∀xP(x) ∨ R)
∃x(P(x) ∧ R) ≡ (∃xP(x) ∧ R)
∀x(P(x) → R) ≡ (∃xP(x) → R)
∃x(P(x) → R) → (∀xP(x) → R)
∀x(R → Q(x)) ≡ (R → ∀xQ(x))
∃x(R → Q(x)) → (R → ∃xQ(x))
∀xR ← R
∃xR → R
The following formulas are not valid.
A B counterexample
∃x(P(x) ∧ Q(x)) ← (∃xP(x) ∧ ∃xQ(x)) D = {a, b}, M = {P(a), Q(b)}
∀x(P(x) ∨ Q(x)) → (∀xP(x) ∨ ∀xQ(x)) D = {a, b}, M = {P(a), Q(b)}
∀x(P(x) → Q(x)) → (∃xP(x) → ∀xQ(x)) D = {a, b}, M = {P(a), Q(a)}
∀x∃yT(x,y) → ∃y∀xT(x,y) D = {a, b}, M = {T(a,b), T(b,a)}
∃x(P(x) → R) ← (∀xP(x) → R) D = Ø, M = {R}
∃x(R → Q(x)) ← (R → ∃xQ(x)) D = Ø, M = Ø
∀xR → R D = Ø, M = Ø
∃xR ← R D = Ø, M = {R}
Note: if empty domains are not allowed, then the last four implications are in fact valid.
Best Answer
In principle, we could make up arbitrarily many distinct quantifiers. $\forall x\colon \phi(x)$ and $\exists x\colon \phi(x)$ make certain statements about the class $C:=\{\,x\mid \phi(x)\,\}$: One says that $C$ is the all-class (or that its complement is empty), the other says that $C$ is not empty. And $\exists!$ say that $C$ is a singleton. We could introduce quantifiers for quite arbitrary statements about $C$, but the most natural would perhaps be about cardinality of $C$: Stating that $C$ or its complement has at least $n$, at most $n$, more than $n$, less than $n$, or exactly $n$ elements for a fixed finite or infinitie cardinality $n$ come to mind. Of these, those with finite cardinalities are readily constructed from $\exists$ and $\forall$ (just like we do with $\exists!$).
However, any quantifiers about infinite cardinalities are problematic in a first order theory (i.e., when we can only "speak" about the objects of our universe, but not about sets of objects of our universe): There are no suitable rules of inference accompanying them. A proof of $\exists x,\phi(x)$ can consist of constructing a sing $a$ with $\phi(a)$. What should a proof of $\exists^\infty x,\phi(x)$ with $\exists^\infty$ intended to mean "there are infinitely many" look like? By exhibiting infinitely many $a_n$ with $\phi(a_n)$? You cannot do that in a first-order theory per se.
Nevertheless, outside first order formalism, some such quantifiers are very common: A very useful one is "almost all" which in the context of natural numbers is understood to mean "all but finitely many". "Almost all" $n$ have property $\phi$ is then a shortcut for $\exists n_0\in\Bbb N\colon \forall n\in \Bbb N\colon n>n_0\to \phi(n)$, a construct very common in the introduction of limits. Note how the very properties of the set of natural numbers allow us to express "all but finitely many". (In other contexts, such as measure theory, we use "almost all" with a different meaning, namely "up to a set of measure $0$")