If $S$ is any constant or variable then '$S$ is a set' is a logical proposition? I have a textbook that claims the following:

Set theory involves the introduction of the new phrase is a set and new symbols $\{ ...\mid ... \}$ and $\in$ defined by
(Set$1$) If $S$ is any constant or variable then '$S$ is a set' is a logical proposition.
(Set$2$) If $P(x)$ is a logical proposition involving a variable $x$ then $\{ x \mid P(x) \}$ is a set.
(Set$3$) If $S$ is a set and $a$ is any constant or variable then $a \in S$ is a logical proposition, for which we say $a$ belongs to $S$ or $a$ is a member of $S$, or simply $a$ is in $S$. The negative of this proposition is denoted $a \not \in S$ - said $a$ is not in $S$.

I find the following axiom peculiar:

(Set$1$) If $S$ is any constant or variable then '$S$ is a set' is a logical proposition.

I have never heard of such a thing. How is it reasonable to say that a constant or variable is a set? This makes no sense to me.
 A: What you're dealing with here is called "naive set theory". 
The first 'axiom' is just a definition of terminology. I don't believe it's really an 'axiom', since we're allowed to make statements anyway. The same holds for (Set3), which is just a definition of terminology. The notion of what "$S$ is a set" means can be made more precise using models, but we first need to look at a problem with the second axiom (which could be considered a proper axiom):
Including the second axiom makes the system inconsistent, since we can take the property $P(x)=x\notin x$, and then $S=\{x\mid x\notin x\}$ would be a set. But $S\in S$ and $S\notin S$ are then both provable, so we have a provable contradiction. This problem is commonly known as Russell's paradox. Inconsistency is of course not something you want in an axiomatic system, since from a contradiction anything becomes provable. 
In a proper introduction of set theory, the definition of the axioms of set theory are a lot more careful to avoid those problems, but for day to day purposes you could ignore the technicalities and just assume anything you can 'reasonably' define, is a set. Only taking definitions that do not imply a sets could contain themselves is already a good start of what would be considered 'reasonable'.

In modern, non-naive set theory, as far as we know the axioms are consistent. This means we can work with models to make precise what the meaning of "$S$ is a set" is. An axiom system has at least one model if and only if it is consistent, so we cannot use this approach for naive set theory (since it's inconsistent).
A model of a certain set theory is a collection of objects and an interpretation of the symbols involved with the theory by those objects, such that all the axioms of the theory become valid given that interpretation. A "set" is then just the name of the objects in the model. 
The sentence "$S$ is a set" is a little strange, since it just means that $S$ is an element of your model. However, the model itself is usually not expressible in the language (as it is usually not a set). 
I would interpret "$S$ is a set" as a meta expression to talk about the model, and not a proposition to talk about what is a set within the model. You could express it as the sentence $\exists x(S=x)$, i.e. you say that there is an object in your model that is equal to $S$.
However, we mix meta language and inside language when talking about mathematics all the time, so we have no problem to understand what is meant by "The collection of all sets" (i.e. all elements of the model) or "$S$ is a set" (i.e. $S$ is part of the model). Therefore I would not worry about it too much, or if you do, it's probably best to read an introduction in axiomatic set theory.

In other kinds of set theory, we have objects of different types. Then not everything has to be a set, therefore making it a more useful expression to be able to state which objects are sets, and which are not. 
