Are sets just predicates with syntactic sugar? Do mathematicians agree/accept that "sets are just predicates with syntactic sugar"?
If not, then Why not? I mean, I can translate between $ x \in S $ and $ S(x) $. Will that change the correctness of a proof?
Is there some nuance in university-level math that makes the distinction necessary?
 A: Set theory is a form of logic, and logic is a form of set theory. In my opinion this is made most clear in the form of category theory.
However, I would not expect the "typical" mathematician to be sufficiently fluent in formal logic to deal with this comfortably.
In my opinion, the typical mathematician works in first-order logic on top of (some fragment of) ZFC. Having two "universes" like that is a subtle and confusing thing — maintaining a sharp distinction between "predicate" and "set" is how the typical mathematician keeps them straight.
A: It depends which axioms you are using. In ZF, the predicate $S(x) = \text{True}, \forall x$ does not correspond to a set.
A: It depends on your framework, but in structuralist mathematics, sets and predicates are very different. Here's the low-down:
Idea 0. Sets form a category. Its called $\mathbf{Set}$.
Idea 1. Given an object $X$ of the category $\mathbf{Set}$, we define a new category $\mathrm{Sub}(X)$ as follows:


*

*an object $\mathfrak{a}$ consists of a set $\overline{\mathfrak{a}}$ equipped with an injection $\eta_\mathfrak{a} : \overline{\mathfrak{a}} \rightarrow X$.

*a morphism $g:\mathfrak{a} \rightarrow \mathfrak{b}$ consists of a function $\overline{g} : \overline{\mathfrak{a}} \rightarrow \overline{\mathfrak{b}}$ such that $\eta_{\mathfrak{b}} \circ 
\overline{g} = \eta_{\mathfrak{a}}$.
Its easy to show that $\mathrm{Sub}(X)$ is a thin category. This means we can treat it as a partially-ordered set; $\mathfrak{a} \subseteq \mathfrak{b}$ means there exists a morphism $\mathfrak{a} \rightarrow \mathfrak{b}$, and $\mathfrak{a} = \mathfrak{b}$ means there exists an isomorphism $\mathfrak{a} \rightarrow \mathfrak{b}$, or equivalently, that there is a morphism going each way.
Idea 2. The poset of truthvalues can be defined as $\mathrm{Sub}(1)$, where $1$ is the set with $1$ element.
Idea 3. Given an object $X$ of the category $\mathbf{Set}$, we define a new category $\mathrm{Pred}(X)$ as follows:


*

*an object $P$ is just a function $P : X \rightarrow \mathrm{Sub}(1)$.

*a morphism $g:P \rightarrow Q$ is just a guarantee that for all $x \in X$, we have $P(x) \subseteq Q(x)$.
As before, $\mathrm{Pred}(X)$ also happens to be a thin category, for all sets $X$.
Idea 4. We can view both $\mathrm{Sub}$ and $\mathrm{Pred}$ as functors $\mathbf{Set}^{op} \rightarrow \mathbf{Pos}$.


*

*$\mathrm{Sub}(f)$ is given by taking pullbacks: $$\mathrm{Sub}(f)(\mathfrak{b}) = f^{-1}(\mathfrak{b})$$

*$\mathrm{Pred}(f)$ is given by precomposition: $$\mathrm{Pred}(f)(P) = f \circ P$$


Idea 5. These functors turn out to be naturally isomorphic, thereby legitimizing the whole "subsets and predicates are the same thing" viewpoint.

So in this language, your question becomes:

Are subsets just predicates with syntactic sugar?

Yes, they're pretty much the same thing; once you've got a few definitions and observations under your belt, at least. You can indeed switch between them in that way.
