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My question deals with uniqueness proofs. For example the proof of the uniqueness of the empty set, or the proof of the uniqueness of the identity element in a group.

These proofs are convincing of course, but are they absolutely rigorous?

Is it possible to consider the following objection: when I prove that empty-set- 1 and ( hypothetical ) empty-set-2 are in fact equal, I prove that the total number of empty sets is not equal to 2, but is this the same as showing that the total number of empty sets is equal to 1?

How to make totally explicit the logic that is behind uniqueness proofs?

Is this logic questionable?

Sorry, but the only correct way to fully understand uniqueness is to look at the logical definition. We say that there is a unique object satisfying a property $Q$ iff we can prove the following:

$∃x ( Q(x) ∧ ∀y ( Q(y) ⇒ x=y ) )$.

(There exists an object $x$ such that $x$ satisfies $Q$ and every object that satisfies $Q$ is equal to $x$.)

It turns out that you can prove in pure first-order logic that this is equivalent to:

$∃x ( Q(x) ) ∧ ∀x,y ( Q(x) ∧ Q(y) ⇒ x=y )$.

(Some object satisfies $Q$ and every objects $x,y$ that satisfy $Q$ are equal.)

Hence you may see an author/teacher prove either version to prove:

$∃!x ( Q(x) )$.

(This is the symbolic representation of "There exists a unique $x$ that satisfies $Q$.".)

Your question is a bit too vague for me to be able to tell which version your teacher/textbook is using. But the point is that uniqueness is defined as per above, and not in terms of 'number of' things that satisfy $Q$, because in general that is not well-defined.

When you show that $a = a'$ in the proof of uniqueness, $=$ is an equivalence relation. Hence if you have a collection of $a$s which are "unique" candidates that actually exist, no matter how large (can be more than two), you partition them into equivalence class(es), and there is only one since all of them are equal.

Take the collection of all the different $a$'s satisfying the property of interest. If the size of this collection is at-least two (can be infinite) we can take two distinct $a$'s out of this collection. Let's call them $a_x$ and $a_y$.

By your proof $a_x=a_y$ , which is a contradiction since they are distinct!

Hence, your proof doesn't merely dismiss the possibility of there being two (for instance) null sets, but actually dismisses the possiblity of there being two or more of them.

The immediate conclusion that follows is that there is either one object satisfying the property or none at all. Showing that it is the former and not the latter is where the existence proof comes in.

Yes, because you can sort of proceed in a pseudo-inductive manner ("pseudo-" because it has that flavor, at least to me, but it's not quite induction).

Since you assume nothing special about, say, $emptyset_1, emptyset_2$ and show that $emptyset = emptyset_1 = emptyset_2$, there's no real difference by considering arbitrary indices. In general you could say $emptyset_i = emptyset_j = emptyset$ for all $i,j$ in some indexing set $I$.

Thus, you conclude that there can be no more than one empty set.

This means there's either one or no empty sets. By showing the existence of the empty set, you get that there is exactly one empty set. Thus, you get uniqueness.