Okay, now I think I understand your question.  This is the question I will answer:

* Question: Let $X$ be a connected $4$-dimensional subspace of $SE(3)$ that contains $SO(3)$.  Is it possible for $X \setminus SO(3)$ to be connected?  Disconnected? 

The answer to both questions is yes.  So there is no Jordan separation theorem for $4$-dimensional subspaces of $SE(3)$ containing $SO(3)$.  

Observation 1: As a space, $SE(3)$ is just the cartesian product of $SO(3)$ with $\mathbb R^3$. Explicitly, we will think of $SE(3)$ as the set $SO(3) \times \mathbb R^3$. 

Observation 2: If $X := SO(3) \times \mathbb R$ embeds in $SE(3)$, therefore $SO(3) \times \{0\}$ disconnects it. 

Observation 3: If $X := SO(3) \times S^1$, where $S^1 = \{ x \in \mathbb R^2 : |x|=1\}$, then the map $X \to SO(3) \times \mathbb R^3$ given by $(p,x) \longmapsto (p,x,0)$ is an embedding.  In particular, $X \setminus (SO(3) \times \{1\})$ is connected. 

So the answer to both your questions is yes.  

I'd like to suggest looking at the proof of the generalized Jordan-Brouwer theorem in Guillemin and Pollack, or perhaps in an algebraic topology textbook like Bredon's.  This will give you a very flexible set of tools that will let you know quite generally when you can expect a separation theorem, and when you can't. 

Notice: my answer had nothing to do with the fact that $SO(3)$ has a non-trivial fundamental group, or whether or not it is orientable.  The key part of the construction is that $SO(3)$ has co-dimension *at least* $2$ (And actually co-dimension $3$) in $SE(3)$.