A beginner's question is ok I think, especially since it may confuse non-beginners. Let me try to answer as concretely as possible, for the benefit of beginners who usually prefer to see concrete answers to general ones.

We first consider a simple case. We adopt as our meta-theory ZFC. Consider the small category $\mathcal{C}$ whose objects are $$\mathrm{ob}(\mathcal{C}) = \lbrace \emptyset \mid 2^{\aleph_0} = \aleph_1 \rbrace,$$
and there are no morphisms other than the identity morphisms. Thus, $S$ has a single object $\emptyset$ if the continuum hypothesis holds, and is empty if the continuum hypothesis does not hold. We define a functor $F : \mathcal{C} \to \mathsf{Set}$ by $F(X) = X$ and $F(\mathrm{id}_X) = \mathrm{id}_X$. Now of course the limit of $F$ exists, but the answer as to what the limit is depends on the status of the continuum hypothesis:

- first note that by the law of excluded middle either $2^{\aleph_0} = \aleph_1$ or not,
- if $2^{\aleph_0} = \aleph_1$ then the limit of $F$ is the set $\emptyset$,
- if $2^{\aleph_0} \neq \aleph_1$ then the limit of $F$ is a singleton, e.g., $\lbrace \emptyset \rbrace$.

Ok, this was just an exercise in which we get used to the fact that a thing may exist, but what the thing is may depend on the status of an undecidable sentence. Nevertheless, *it exists*, it is just that the question "Is the limit $\emptyset$ or $\lbrace \emptyset \rbrace$?" is undecidable.

There is a trickier question we can ask, and which you are asking I think, namely, can we conjure up a functor $F$ such that its limit is a set whose very existence itself is independent of ZFC. For example, one might try to come up with an $F$ whose limit is a Suslin line. The answer is that you will fail in such attempts. Why? Because ZFC proves that all small diagrams have limits. If the power of proof does not convince you, then perhaps we should look at one specific example that you listed.

You suggested the category $\mathcal{S}$ of "Suslin objects" (by which I mean dense totally ordered sets without endpoints satisfying CCC) with order-isomorphisms as morphisms. First of all, as stated $\mathcal{S}$ is not small, but we can make it small by limiting the rank of its objects to something like $\omega + 1$ (correct me if my ordinal is too small, I just need an $\alpha$ such that $V_\alpha$ includes isomorphic copies of all possible Suslin objects). The functor $F : \mathcal{S} \to \mathsf{Set}$ is just the underlying-set functor. The trouble here of course is that the question "Does $\mathcal{S}$ contain a Suslin line?" is undecidable, so we might worry that the existence of the limit of $F$ itself is undecidable. But it isn't! The limit exists, and up to isomorphism it is simply the cartesian product of (the underlying sets of) all Suslin objects whose rank does not exceed $\omega + 1$, *whatever they are*. Of course, the question "what does the limit look like?" depends on existence of Suslin lines.

The other examples you listed are of the same nature.

Look, this has nothing to do with category theory. Consider the triangle in the plane whose vertices have coordinates $A = (0,0)$, $B = (0,1)$ and $C = (x,1)$ where $x = 1/2$ if Suslin line exists and $x = 42$ otherwise. The triangle exists, but telling whether it is isosceles is a bit difficult. Your worry is of exactly the same kind, I think.