I believe that the quantifier "Most $x$ have property $P$" is not first-order definable, that is, not definable using universal and existential quantifiers and equality. More precisely, if we restrict our attention to the class of finite $\{P\}$-structures, (where $P$ is a unary predicate symbol), there is no formula of first-order logic that holds precisely of those structures where the cardinality of the set $\{x | P(x)\}$ is strictly greater than its complement $\{x | \neg P(x)\}$. However, I want to see a proof of that statement. It is intuitively plausible to me, but I want a proof that there is no clever way to define that quantifier.
Proof of undefinability of the “Most” quantifier in first-order logic
logicquantifiers
Related Solutions
Note that neither of the two rules you quote are formulated as "$(C\to A(x))\to(C\to\forall x.A(x))$". The middle $\to$ there is not in any of the rules -- they carefully use "if ... then" rather than a symbolic "$\to$", which means that the implication they express is at the level of logic-as-a-formal-game rather than at the level of truth in a particular interpretation. By reading it as $\to$ you have inadvertently made the rules into nonsense.
What the rule tells you is that if the string of symbols $$C\to A(x)$$ is valid in the interpretation you speak about, then the string of symbols $$ C\to\forall x.A(x)$$ is also valid. This is something that holds about particular strings of symbols, some of which symbols happen to be $x$, but it's not something that holds for any particular value of $x$ -- at the levels the rules work at, $x$ is just chalk and it does not have values.
To say that $C\to A(x)$ is valid means that it evaluates to true under every assignment of its free variables. Therefore the claim "$C\to A(x)$ is valid in such-and-such interpretation" does not itself have $x$ as a free variable; it is not a claim that can even in principle be true for some but not all values of $x$ any more than it can be true for some but not all values of $q$.
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.
Best Answer
I think your claim follows from this zero-one law for first-order sentences in finite models.
If $\sigma$ consists of a single unary predicate $P$, then the probability that a random $\sigma$-structure of cardinality $n$ satisfies "most $x$ have property P" approaches $\frac12$ as $n\to\infty$, so this sentence cannot be first-order.