I cannot find the proof you are looking for in the book by D. E. Taylor
that ego suggested in the comments either. But you can use the techniques
of the section "Reflections and the Strong Exchange Condition" on pages
94-96 for a proof (my answer is independent of the book, for connections
see the comment at the end).
In your case you have the set $R = \{(1\;-1)\} \stackrel{.}{\cup}
\{(i\;i+1)(-i\;-i-1) \mid i \in [n-1]\}$ of involutions generating the
Coxeter group $W = \langle R\rangle$, which is a subgroup of the symmetric
group on $[n] \cup -[n]$ (using the notation $[n] := \{1, \dots, n\}$).
For the set $T = \{wrw^{-1} \mid w \in W, r \in R\}$ of all conjugates of
the generating involutions in $R$ (defined in equation (9.20) of the book) you
should be able to check
$T = T_1 \stackrel{.}{\cup} T_2 \stackrel{.}{\cup} T_3$ for
$$T_1 := \{(i\;-i)\mid i\in[n]\}$$
$$T_2 := \{(i\;j)(-i\;-j) \mid i, j \in [n], i<j\}$$
$$T_3 := \{(i\;-j)(-i\;j) \mid i, j \in [n], i<j\}$$
$T$ consists of two different conjugacy classes:
$T_1$ are the conjugates of $(1\;-1)$), and $T_2\stackrel{.}{\cup}T_3$ the
conjugates of all the other elements of $R$.
$W$ acts on $T$ by conjugation, which induces an action on the power set
$\mathcal{P}(T)$ of $T$. With the symmetric difference as addition, the
power set $\mathcal{P}(T)$ becomes an elementary abelian $2$-group, on
which $W$ acts. Define
$D'(w) = D'_1(w) \stackrel{.}{\cup} D'_2(w) \stackrel{.}{\cup} D'_3(w)$
for $w \in W$ with
$$D'_1(w) = \{(i\;-i) \in T_1 \mid i\in [n] \mbox{ and } w(i)<0\}$$
$$D'_2(w) = \{(i\;j)(-i\;-j) \in T_2 \mid i, j\in [n], i<j\mbox{ and }
w(i)>w(j)\}$$
$$D'_3(w) = \{(i\;-j)(-i\;j) \in T_3 \mid i, j\in [n], i<j\mbox{ and }
w(i)+w(j) < 0\}$$
The cardinality of $D'(w)$ is just inv$(w)$ from your question.
Claim:
(a) $D'(r) = \{r\}$ for $r \in R$
(b) $D'(w_1w_2) = w_2^{-1}D'(w_1)w_2+D'(w_2)$ for $w_1, w_2 \in W$
The first statement is easily verified, we show the second one:
For this, observe that an element $(i\;-i)$ of the conjugacy class $T_1$
is contained in $D'(w)$ if and only if $\frac{w(i)}{i}<0$ for
$i \in [n]\cup-[n]$.
Now $(i\;-i) \in D'(w_1w_2)$ is the same as $0 > \frac{w_1(w_2(i))}{i} =
\frac{w_1(w_2(i))}{w_2(i)}\cdot \frac{w_2(i)}{i}$, which in turn is
equivalent to either $ D'(w_1) \ni (w_2(i)\;-w_2(i)) = w_2(i\;-i)w_2^{-1}$
or $(i\;-i) \in D'(w_2)$, i.e., $(i\;-i) \in w_2^{-1}D'(w_1)w_2+D'(w_2)$.
An element $(i\;j)(-i\;-j)$ of the conjugacy class $T_2\cup T_3$, where
$i, j\in [n]\cup -[n]$, $i\ne j$ and $i\ne -j$, is contained in $D'(w)$
if and only if $\frac{w(j)-w(i)}{j-i}<0$ (check this by considering the
two cases $i\cdot j>0$ and $i\cdot j<0$ and by considering what happens
if you replace $i$ and $j$ by $-i$ and $-j$).
So $(i\;j)(-i\;-j) \in D'(w_1w_2)$ is equivalent to
$0 > \frac{w_1(w_2(j))-w_1(w_2(i))}{j-i} =
\frac{w_1(w_2(j))-w_1(w_2(i))}{w_2(j)-w_2(i)}\cdot
\frac{w_2(j)-w_2(i)}{j-i}$, which is the same as
$(i\;j)(-i\;-j) \in w_2^{-1}D'(w_1)w_2+D'(w_2)$.
Having proven the claim, one can easily deduce the formula for the length by
induction:
Because of (a) we may assume that $l(w) > 1$, and write $w = s\cdot v\cdot t$
with $s, t \in R$ and $l(v) = l(w)-2$. Per induction
$|D'(v)|+1 = l(v)+1 = l(v\cdot t) = |D'(v\cdot t)| \stackrel{(b)}{=}
|tD'(v)t + D'(t)| \stackrel{(a)}{=} |tD'(v)t + \{t\}|$, i.e., $t \not\in D'(v)$.
As $v\ne w = svt$ we get $t \ne v^{-1}sv$, and hence
$t\not\in \{v^{-1}sv\} + D'(v) \stackrel{(a)}{=} v^{-1}D'(s)v + D'(v)
\stackrel{(b)}{=} D'(s\cdot v)$.
It follows $D'(w) \stackrel{(b)}{=} tD'(s\cdot v)t \stackrel{.}{\cup} \{t\}$ and
therefore the induction step.
The claim is a variant of (9.22) in Taylor's book:
For $D(w) := wD'(w)w^{-1}$ you get $D(r) = r\{r\}r = \{r\}$ as $r$ has
order 2.
Also $D(w_1w_2) = w_1D'(w_1)w_1^{-1}+w_1w_2D'(w_2)w_2^{-1}w_1^{-1} =
D(w_1)+w_1D(w_2)w_1^{-1}$ showing that D fulfills (9.22).
This condition on $D$ is quite powerful. Taylor uses it to derive the strong
exchange property (with Corollary 9.26 implying the formula for the length), and
that it is equivalent to $(W, R)$ being a Coxeter system.
If $n=3$ then there might be more subspaces, for example the span of $(1,\omega,\omega^2)$ where $\omega=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$.
If $n>3$ then the subspaces you listed are indeed all the subspaces. Let $e_1,...,e_n$ be the standard basis vectors of $\mathbb{C^n}$. We can define an action of $A_n$ on $\{e_1,...,e_n\}$ by $\sigma.e_i=e_{\sigma(i)}$. Just like any other action, it induces a representation of $A_n$. Formally, the representation is defined as $\pi:A_n\to GL(\mathbb{C^n})$ where $\pi(\sigma)(a_1e_1+...+a_ne_n)=a_1e_{\sigma(1)}+...+a_ne_{\sigma(n)}$. So we are actually looking for the invariant subspaces of this representation.
We can obviously define the invariant subspaces $V_1$ and $V_2$ like you did in the question. The subspace $V_1$ is one dimensional, thus irreducible. Also, since $n>3$ the action of $A_n$ on $\{e_1,...,e_n\}$ is $2$-transitive. (can you prove it?). This implies that if $\chi$ is the character of $\pi$ and $\chi_1$ is the character of the trivial representation then $\chi-\chi_1$ is an irreducible character. Since $\mathbb{C^n}=V_1\oplus V_2$ it follows that $\chi-\chi_1$ is exactly the character of $V_2$, so $V_2$ is also irreducible.
Now the result follows. Take a non-trivial invariant subspace $U$. It has an invariant complement $W$, which also has to be non-trivial. By Maschke's theorem you can decompose both into a direct sum of irreducible subspaces. But such a decomposition of $\mathbb{C^n}$ is unique, it is exactly $V_1\oplus V_2$. So we have no choice, $U$ must be either $V_1$ or $V_2$.
Best Answer
I think that what you are calling a "sensible action" would more usually be called an "action by semigroup automorphisms". Actually, in most contexts, when a group acts on a semigroup this property would be implicitly assumed (or else why mention the semigroup structure at all?).
Anyway, at the level of general semigroups, there is probably not much to be said. Given any set $S$ and a choice of a point $0 \in S$, defining $xy=0$ for all $x,y \in S$ makes $S$ into a semigroup. A bijection $\phi : S \to S$ will be a semigroup automorphism (i.e. will satisfy $\phi(xy)=\phi(x)\phi(y)$ for all $x,y\in S$) if and only if $\phi(0)=0$. So the semigroup automorphisms of $S$ are just the permutations of $S$ which fix the point $0$.
Obviously there is not much of a difference between general group actions on sets, and group actions on pointed sets which preserve the base point. There is a straightforward recipe for going between the two (add/remove the basepoint).
So, one response to your question "what properties do group actions on semigroups have that group actions on sets lack?" would be "not much". Without restricting attention to some special type of semigroups, basically any group action on a set can be put into this context.
Another comment: given a set $X$, you can still form a vector space $F[X]$, the vector space having the elements of $X$ as a basis. It doesn't have an interesting algebra structure, like it would in the semigroup case, but let's not worry about that. Now, if $G$ is acting on the set $X$, it also acts on the vector space $F[X]$ by linear transformations (namely ones that permute the basis). So you still get a permutation representation from a group action on a set.