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.
The standard notation in logic would be $\exists^\infty$.
The exclamation mark ! is used to indicate uniqueness, $\exists^{!n} x\,\phi(x)$ being "there are exactly $n$ distinct elements $x$ such that $\phi(x)$". So, the standard reading of $\exists^{!\infty}x\,\phi(x)$ would be "there are exactly infinitely many $x$ such that..." which is awkward, as it is not very exact to simply say "infinitely many", since there are many options here. And if you are in a setting where the universe of discourse is countably infinite, for instance, then the "exactly" part is superfluous anyway.
You are correct that $\exists^{\infty}x\,\phi(x)$ is the same as $\forall n\,\exists^{\ge n}x\,\phi(x)$. This is the case regardless of whether you are in a situation where the axiom of choice holds. I mention this because using choice, $\exists^{\infty}x\,\phi(x)$ is equivalent to $Q_{\aleph_0}x\,\phi(x)$, where $\aleph_0=|\mathbb N|$ and, for a cardinal $\kappa$, $Q_\kappa x\,\phi(x)$ means that there are at least $\kappa$ distinct values of $x$ such that ... (The $Q_\kappa$ are called cardinality quantifiers in the literature.) However, if choice fails, a set may be infinite without containing a countably infinite subset.
(And just to avoid confusion, let me add the remark mentioned in comments: Although each $\exists^{\ge n}$ is first-order definable as shown in the question, $\exists^\infty$ is a genuinely new quantifier, meaning that it is not first-order definable by a formula. Otherwise, its negation ("there are only finitely many") would also be first-order definable, and an easy compactness argument gives us a contradiction: the theory $$\{\lnot\exists^\infty x\,(x=x)\land\exists^{\ge n}x\,(x=x)\mid n\in\mathbb N\}$$ is inconsistent but any finite subset is consistent.)
Best Answer
If this doesn't completely answer your question, I can expand it later if you ask a followup in a comment. The simple answer is just that there's a difference between letting a formula have infinite length and letting the interpretation (and truth-value) of a formula depend on infinitely many assignments. You make a clear split between (i) the formal language and its deductive system as just mechanical manipulations of meaningless strings of symbols, (ii) the structures that you interpret the language to be talking about, and (iii) the way that the interpretations (mappings from (i) to (ii)) themselves work. Very generally/vaguely, restricting things to be finite should keep things simpler. This seems reasonable, yes? So there might be advantages to confining the infinite bits to one or two of the above steps. It turns out that this does make a difference, and this is one thing that introducing quantifiers allows you to do.
To answer your questions more directly:
(1) Yes, in some sense, but where these infinite things are and how precisely they are infinite (i.e., what their structure is) is another matter. In FOL, they are relegated to the metalanguage and metatheory.
(2) This seems like a reasonable way to think of things, but it depends on the details of what you mean. De Morgan's laws are about relationships between the logical operators, and the operators are exactly the same in propositional and predicate/first-order logic. De Morgan's laws do hold in FOL -- they're exactly the same as in propositional logic. If you want to point out a relationship among the concepts of negation, conjunction, and disjunction that we use in the different steps above (in the object language and metalanguage) and how they relate to quantification, then yes, you seem to have the right idea. The reason that the existential and universal quantifiers are related in the way that you point out is because of how negation, conjunction, and disjunction are related in the metatheory. But you might be able to trace this idea back to Aristotle and company.
Cheers, Rachel
Edit to answer comment: Yes, the distinction between finitary and infinitary logics usually has to do with the formal language and formal deduction part. You can allow formulas or proofs (or both) of infinite length.
The same distinction between formal, syntactic types of things and meaningful, semantic types of things is also applied to propositional logic. It is not always brought to everyone's attention when studying propositional logic because it's a simpler system, and the pieces are related in a way that makes the distinction not of much consequence. The thing to note is that the same steps are happening in propositional logic, but humans don't need to be told how to do them because using natural language has already taught them how to do this mapping between form and meaning. It's more like natural language has made the steps invisible, and they need to be pointed out. Well, also, people don't consciously know how they use natural language, and that is where the work is -- to make these things explicit.
If you're interested in logic and a bit of model theory, two very good books are Hodges' Logic and Machover's Set theory, logic, and their limitations. Hodges is incredibly funny and insightful, and Machover is very good about pointing out clearly and thoroughly the relationships just touched on.