I grant that logical equivalents not containing 'exist(s)' or cognates can be supplied for all singular and general existentials. Thus, 'Socrates exists' can be translated, salva veritate, as 'Something is identical to Socrates,' or, in canonical notation, '(∃x)(x = Socrates).' Accordingly,
Socrates exists =df (∃x)(x = Socrates).
But if the definiens preserves the truth of the definiendum, then the definiendum must be true, hence must be meaningful, in which case first-level uses of 'exist(s)' must be meaningful. Pace Russell, 'Socrates exists' is nothing like 'Socrates is numerous.'
What's more, the definiendum is prior in the order of understanding to the definiens. If I didn't already understand 'Socrates exists,' then I would not be able to understand '(∃x)(x = Socrates).' You couldn't teach me the Quinean translation if I didn't already understand the sentence to be translated.
One conclusion we can draw from this is that if 'exist(s)' is univocal across general and singular existentials, then existence cannot be instantiation. For the left-hand side of the definition does not make an instantiation claim. It is simply nonsense to say of an individual that it is instantiated. And if the right-hand side makes an instantiation claim, then we need those creatures of darkness, haecceity-properties.
But we don't have to give the RHS a Fressellian reading; we can give it a Quinean-Inwagenian reading. (We could call this the 'Van' reading.) Accordingly: There exists an x such that x = Socrates. On the Van reading, in stark contrast to the Fressellian reading, 'exist(s)' can be construed as a first-level predicate, as synonymous to the predicate 'is identical to something.' Accordingly:
y exists =df(∃x)(x = y).
On the reasonable assumptions that (i) 'exist(s)' is an admissible first-level predicate and that (ii) there are no nonexistent objects, this last definition is unobjectionable. If Tom exists, then there exists an object to which he is identical. And if there exists an object to which Tom is identical, then Tom exists. No doubt!
The interesting question, however, is whether any of this affords aid and comfort to the thin theory. Well, what exactly is the thin theory? It is the theory that existence is exhaustively understandable in purely logical, indeed purely syntactical, terms. The thin theory is a deflationary theory that aims to eliminate existence as a metaphysical topic. It aims to supplant the metaphysics of existence (of whatever stripe: Thomist, Heideggerian, etc.) with the sober logic of 'exist(s).' The aim of the thin theory is to show that there is no sense in which existence is a non-logical property of individuals. The aim is to be able to consign all those tomes of metaphysical rubbish to the flames with a good conscience.
Now glance back at the definition. Every mark on the RHS is a bit of logical syntax. Ignoring the parentheses which in this instance can be dropped, we have the backwards-E, two bound occurrences of the variable 'x,' a free occurrence of the variable 'y,' and the sign for identity. There are no non-logical expressions such as 'Socrates' or 'philosopher.' On the LHS, however, we find 'exists' which is not obviously a logical expression. Indeed, I claim that it is not a logical expression like 'some' or 'all' or 'not.' It is a 'content' expression. What could be more important and contentful than a thing's existing? If it didn't exist it would be nothing and couldn't have properties or stand in relations.
Surely my sheer be-ing is my most impressive 'feature.' "To be or not to be, that is the question."
Since there is content on the LHS there has to be content on the RHS. But how did it get there, given that every expression on the RHS is just a bit of syntax? In only one way: the domain of the bound variables is a domain of existents. But now it should be clear that the definition gives us no deflationary account of existence. What it does is presuppose existence by presupposing that the domain of quantification is a domain of existents. Existence is that which existents have in common and in virtue of which they exist.
In short, I have no objection to the definition read in the 'Van' as opposed to the 'Fressellian' way. It is perfectly trivial! My point, however, is that it gives no aid and comfort to the thin theory. A decent thin theory would have to show how we can dispence with existence entirely by eliminating it in favor of purely logical concepts. But that is precisely what we cannot do given that the domain of quantification is a domain of existents. (Of course, if the domain were populated by Meinongian nonexistent objects, then the definition would be false).