Your reasoning isn't correct, I'm afraid, but all your conclusions are.
Yes, the orbit of a vertex has the eight vertices in it.
Yes, the stabilizer of a vertex has three rotations in it.
Two of the rotations you listed are the identity rotation, because rotating by $2\pi$ leaves the cube exactly in it's original position. If that didn't count as the identity, you would have infinitely many symmetries, one for each full turn cockwise or anticlockwise, but no, we don't consider the route, we consider the transformation from start position to end position, and since they're the same, it's the identity rotation.
Why are there three elements in the stabilizer for the vertex? ![[Diagonal axis rotation]](https://i.stack.imgur.com/bui0h.gif)
When you rotate a cube about its longest diagonal axis, there are three faces that are permuted cyclically (in the picture, the centre of the red, yellow and blue faces are an orbit of three, and the black, white and green faces are another). This stabilizer is a cyclic group of order three, because you can do that rotation three times before getting back to where you started, as you can visualize by watching the picture.
Hence the stabilizer of a vertex under rotations of the cube consists of three elements: 1. the identity rotation (by $0$ or $2\pi$ or $-24\pi$, it's all the same symmetry), 2. rotation about the long diagonal axis by $2\pi/3$ and 3. by twice that.
The main point : all your subgroups are rather small and
generated by a single rotation.
There are at least 24 elements in $G$ :
The identity ${\sf id}$,
Rotations around axes through the middles of one of
3 pairs of opposing faces and with a rotation of
$\frac{k\pi}{2}(1\leq k \leq 3)$ :there are $3\times 3=9$ such rotations,
and we denote them by $OF(<name\ of\ face>,<angle>)$
Half-turns around axes through the middles of one of
6 pairs of opposing edges : there are $6$ such rotations, and we denote them by $HT(<name\ of\ opposite\ edges>)$.
Rotations with axis $18,26,37$ or $45$, and angle
$\frac{2k\pi}{3}(1\leq k \leq 2)$ :there are $2\times 4=8$ such rotations,
and we denote them by $R(<name\ of\ axis>,<angle>)$.
Since there are at most $24$ orientation-preserving of the cube
(see here), we see that $G$ consists exactly of the $24$ elements
enumerated above.
Let $G_v$ denote the subgroup of $G$ fixing the vertex $1$, $G_e$ the subgroup
fixing the edge $12$, and $G_f$ the subgroup fixing the face $1234$.
By the enumeration above, we have :
$$
\begin{array}{lcl}
G_v &=& \lbrace {\sf id};R(18,\frac{2k\pi}{3}) (1 \leq k \leq 3) \rbrace \\
G_e &=& \lbrace {\sf id};HT(12,78) \rbrace \\
G_f &=& \lbrace {\sf id};OF(1234,\frac{k\pi}{2}) (1 \leq k \leq 3) \rbrace
\end{array}
$$
It is easy then to deduce the decompositions into orbits :
$$
\begin{array}{|l|l|}
\hline
\text{Subgroup} & G_v \\
\hline
\text{Orbits in } V &
[1] [2,3,5] [4,6,7] [8] \\
\hline
\text{Orbits in } E &
[12,13,15] [24,37,56] [26,34,57] [48,68,78] \\
\hline
\text{Orbits in } F &
[1234,1256,1357] [2468,3478,5678] \\
\hline
\end{array}
$$
$$
\begin{array}{|l|l|}
\hline
\text{Subgroup} & G_e \\
\hline
\text{Orbits in } V &
[1,2] [3,6] [4,5] [7,8] \\
\hline
\text{Orbits in } E &
[12] [13,26] [15,24] [34,56] [37,68] [48,57] [78] \\
\hline
\text{Orbits in } F &
[1234,1256] [1357,2468] [3478,5678] \\
\hline
\end{array}
$$
$$
\begin{array}{|l|l|}
\hline
\text{Subgroup} & G_f \\
\hline
\text{Orbits in } V &
[1,2,3,4] [5,6,7,8] \\
\hline
\text{Orbits in } E &
[12, 13, 24, 34] [15, 26, 37, 48] [56, 57, 68, 78] \\
\hline
\text{Orbits in } F &
[1234] [1256,1357,2468,3478] [5678] \\
\hline
\end{array}
$$
Best Answer
It's enough to show that any edge can be rotated into one of the neighbouring edges; by composing such operations, you can move any edge to any other edge. To rotate an edge into a neighbouring edge, rotate through $2\pi/3$ about an axis through the vertex they share.
For an edge to be rotated into itself, the axis has to pass through the centre of the edge and the angle has to be $\pi$; this is the only kind of rotation that leaves a line segment invariant, other than a rotation about an axis along the line segment, which isn't an option in this case.