For completeness, this is the definition of a functor I am working with:
My textbook says that
A functor between sets $S$ and $T$ is a function from $S$ to $T$.
I tried proving this fact. To keep it short, I deduced that all other properties of these kinds of functors (i.e., functors being functions between sets $S$ and $T$) follow from $F_1(\mathsf{1}_A) = \mathsf{1}_{F_0A}$. Because all arrows in $T$ are identities, it surely follows that $F_1(\mathsf{1}_A) = \mathsf{1}_{B}$, but here I get stuck. Without additional information, it seems that the condition $F_1(\mathsf{1}_A) = \mathsf{1}_{F_0A}$ could fail, because $F_1$ could map some $f$ to $Ff : B \to B$, but because there are no constraints on $F_0$, it could be the case that $F_0A \neq B$.
I've gone through the definition carefully and I failed to find an additional premise which would constrain $F_0$ as needed. It seems that the crucial question here is what exactly does it mean for a functor to be a function between sets $S$ and $T$, viewed as categories. After all, a functor has a richer structure than a simple function between sets, being a pair of functions $(F_0, F_1)$.
Thank you for your help. This question is especially important, as I cannot proceed in studying category theory without a clear understanding of the structure of functors.
Addendum. $\mathcal{C}_0$ denotes the class of the objects of a category $\mathcal{C}$, and $\mathcal{C}_1$ the class of its morphisms/arrows.
All other properties of functors do not follow from $F_1(1_A)=1_{F_0A}$, at least not in general. However, if you are viewing sets as so-called "discrete" categories, then what you have said is true, because the only arrow $A\to A'$ in $S$ will just be an identity arrow $A\to A$.
So long as $F_1(1_A)$ is an arrow $F_0A\to F_0A$ (i.e. has correct domain and codomain) then it equals $1_{F_0A}$ by uniqueness, as you say. But this is always true by definition of a functor - you are worrying in the wrong direction, I think? If you are given a functor $F$, then you already know $F_1(1_A)=1_{F_0A}$, you don't need to prove that. If you are constructing a functor $F$, then you are in control of $F_0$ and $F_1$, and $F_1f$ is always forced (for any arrow $f$) to have domain $F_0A$ and codomain $F_0A'$ if $f$'s domain and codomain are $A,A'$ respectively. So there is a constraint, you are forced by law to make that choice. Again, it is up to you to construct $F_1$, but there is actually only one thing you can do.
The thing about discrete categories is that if $S$ is a discrete category and $C$ is any other category, and if you are given $F_0:S_0\to C_0$, there is one and only one way to extend $F_0$ to a functor. This is for exactly the reasons mentioned: $F_1$ must have the right domains and codomains and the only arrows you need to define $F_1$ for are identity arrows, which are obliged to be sent to identity arrows by law. Hence, one and only one way to do it.
This realisation is really the point of the exercise, so think on it.
If you have $F_0,F_1$ given and want some $f:S\to T$, then all you need to do is find, for every point $s\in S$, a point $t\in T$ which you can call $f(s)$. If you want to do this in a way related to $F_0,F_1$, consider that $s$ is an object of the category $S$, so $F_0(s)$ is an object of the category $T$ and can be understood as an element of $t$. So "what it means for a functor to be a function..." is that any functor "descends" to a function $S_0\to T_0$, in this case $S=S_0,T=T_0$, so your desired $f:S\to T$ is just simply $F_0$. You can ignore the data of $F_1$ because, as said, $F_1$ is uniquely determined by $F_0$ (it adds no new information).
Going the other way, any $f:S\to T$ creates an $F_0:S_0\to T_0$, which creates an $F_1:S_1\to T_1$ in a unique way, which creates your functor $F$.