What you call a list is formally known as sequence. There was a question which symbol is for sequence concatenation. Unfortunately there is no accepted answer. Symbols ⋅, ⌒ (commentator actually used u2322, "frown" symbol but it's resisting my attempt to copy it) and ∥ are mentioned in comments.
According the Wikipedia article ∥ is an operator for concatenation of numbers (doesn't specify which set of numbers, probably ℕ) but doesn't say much about sequences. The same symbol is in my opinion more commonly used for parallelism so it may confuse the reader.
I haven't seen ⌒symbol before but commentators agree about it.
It depends on context, but I would say that you are unlikely to cause confusion if you write $A \sqcup B$ for the union of $A$ and $B$, where $A$ and $B$ are disjoint subsets of a larger space. For example, it is a fun fact about the $p$-adic numbers (denoted by $\mathbb{Q}_p$, where $p$ is some fixed prime number) that the unit ball in $\mathbb{Q}_p$ is composed of precisely $p$ congruent balls of radius $\frac{1}{p}$. Indeed, we could reasonably write
$$ B(0,1) = \bigsqcup_{j=0}^{p-1} B(j, \tfrac{1}{p}).$$
Formally, as is noted in the question, the set $A\sqcup B$ is not the same thing as the set $A \cup B$. On the one hand, the union of two sets is defined to be
$$A \cup B := \{ x : x\in A \lor x\in B \},$$
where $A$ and $B$ are subsets of some (thus far unspecified) universal set.
On the other hand,
$$A \sqcup B := (A\times\{0\}) \cup (B\times\{1\}), $$
where $X\times Y$ denotes the Cartesian product and $\cup$ is the union in the first sense. Note that the choice of $\{0\}$ and $\{1\}$ is arbitrary. Really, we just need two distinct elements: one to pair with elements of $A$, and another to pair with elements of $B$.
However, if $A, B \subseteq \mathscr{U}$ and $A \cap B = \emptyset$ (that is, if $A$ and $B$ are disjoint), then there is a correspondence between the union ($\cup$) and the disjoint union ($\sqcup$): define the function $f : A\sqcup B \to A\cup B$ by
$$ f(x,k) := x $$
It is not too hard to see that this function is surjective: if $y \in A \cup B$, then either $y \in A$ and so $(y,0) \in A\sqcup B$ is mapped to $y$ by $f$, or $y \in B$ and so $(y,1) \in A\sqcup B$ is mapped to $y$ by $f$. Injectivity is similarly simple to show.
From a purely set-theoretic point of view, the two sets are isomorphic, i.e.
$$ A \cup B \equiv A \sqcup B. $$
With a bit more work, we can show that this same map preserves other structures (under the right hypotheses). For example, if $A$ and $B$ are subsets of a topological space, then there is a natural topology on the set $A \sqcup B$ such that the disjoint union is homeomorphic to the union.
Best Answer
Just to summarize the comments, the proper notation would be $ \{x\} \cup S,$ since $x \cup S$ implies that $x$ is just a notation for some set that could possibly contain multiple elements.