Expanding out my comment into a full answer. There are two ideas of complexification of Lie groups that are used. Because of the ambiguity and technical awkwardness we tend to sweep it under the rug and talk about complexified Lie algebras and their Lie groups rather than complexified Lie groups.
The first idea is perhaps the obvious one. A complexification of a real Lie group $G$ is a complex Lie group $G_\mathbb{C}$ which contains $G$ as a real Lie subgroup and whose Lie algebra $\mathfrak{g}_\mathbb{C}$ is a complexification of the Lie algebra of $G$. As an example of this $SL_n(\mathbb{C})$ is the complexification $SL_n(\mathbb{R})$.
See here for more detail: https://encyclopediaofmath.org/wiki/Complexification_of_a_Lie_group
This works quite well in many cases and nicely interfaces with the Lie group/algebra correspondence. However it has a major flaw. Not every real Lie group has a complexification even among the semisimple groups. As I understand having a complexification in this sense is equivalent to being linear (so if you're into linear algebraic groups you can stop here). In particular the universal cover of $SL_n(\mathbb{R})$ has no complexification in this sense
To fill in the gaps we define the universal complexification. This is the definition that Wikipedia uses https://en.wikipedia.org/wiki/Complexification_(Lie_group). This one is defined by a universal property. The complexification of $G$ is a complex Lie group $G_\mathbb{C}$ together with a continuous homomorphism $\varphi:G \to G_\mathbb{C}$ such that for any homomorphism to a complex Lie group $f:G\to H$ there is a unique complex analytic homomorphism $F: G_\mathbb{C}\to H$ such that $ f= F \circ \varphi$.
Constructively, we can build this as follows. Take the simply connected group $\tilde{G}$ covering $G$ and the natural homomorphism $\Phi:\tilde{G}\to \tilde{G}_\mathbb{C}$ from this into the simply connected complex group (which we find by complexifying the Lie algebra). The natural projection $\pi:\tilde{G}\to G$ has kernel equal to the fundamental group of $G$. Now we define the universal complexification as $\tilde{G}_\mathbb{C}/N$ where $N$ is the smallest normal subgroup which contains $\Phi(\ker \pi)$.
As you can see the second is much more complicated and can even produce a group whose Lie algebra isn't even the complexification of the original group's but a quotient of it. One neat thing is the universal complexification of a simply connected group is exactly the corresponding simply connected complex group.
I am not sure about the etiquette for answering cross-posted questions. I also posted this answer on MO. If it is not appropriate to post it here, then I am happy to delete it. See also my answer on MO discussing certain special soluble groups, and @YCor's more general answer on MO that handles the general soluble case.
A connected, soluble nilpotent Lie group $G$ has no maximal proper, closed subgroup. (Thanks to @YCor for pointing out on MathOverflow, with a counterexample, that my original claim was incorrect.)
We proceed by induction on the dimension of $G$. If the dimension is $0$, then $G$ is trivial, and we are done; so suppose that the dimension is positive, and hence that $\operatorname Z(G)^\circ$ is a positive-dimensional subgroup of $G$.
Suppose that $H$ is a maximal proper, closed subgroup of $G$. If $H$ contains $\operatorname Z(G)^\circ$, then $H/{\operatorname Z(G)}^\circ$ is a maximal proper, closed subgroup of $G/{\operatorname Z(G)}^\circ$, which is a contradiction. Therefore, $H$ is a proper subgroup of $H\cdot\operatorname Z(G)^\circ$. By maximality, we have that $H\cdot\operatorname Z(G)^\circ$ is dense in $G$. (It seems plausible that $H\cdot\operatorname Z(G)^\circ$ is already closed, but I do not know how to prove it.) Thus, for every pair of elements $g, g' \in G$, we have sequences $((h_n, z_n))_n$ and $((h_n', z_n'))_n$ in $H \times \operatorname Z(G)^\circ$ whose images under the multiplication map $H \times \operatorname Z(G)^\circ \to G$ converge to $g$ and $g'$. Then the sequence $([h_n z_n, h_n'z_n^{\prime\,{-1}}])_n$ equals $([h_n, h_n'])_n$, hence lies in $H$; and converges to $[g, g']$, which therefore belongs to $H$. Then $H$ contains the derived subgroup of $G$, which, as you have observed, is a contradiction.
Best Answer
For each of the three points, I can provide a partial, yet incomplete, answer.
Regarding 1., the following link states that $R \cap S = \{e\}$ (and $G$ is a semidirect product of $S$ and $R$) if $G$ is simply connected (without any other assumptions):
https://www.encyclopediaofmath.org/index.php/Lie_group
Regarding 2., the commutator is a characteristic subgroup, so the answer is yes if $G$ is normal in $PSL_n(\mathbb C)$.
Regarding 3., note that in the compact case, the complex Lie group associated to the complexification of the Lie algebra is the complexification of $G$, so by the universal property of the latter, we get a morphism $G_{\mathbb C} \to PSL_n(\mathbb C)$. The image of $\pi$ has a complexification in $PSL_n(\mathbb C)$ and the inverse of $\pi$ may be extended to it via the same universal property. A third application of that universal property shows that injectivity is preserved.