EDIT We show that the answer to the OP's question is yes. Thanks to Will Sawin for his comments.

I use the notation of [Hartshorne, Algebraic Geometry, Chapter V Section 2].

Since $S$ is a ruled surface, there exists a section $C_0$ of minimal self-intersection; set $C_0^2 = -e$. If we write $S=\mathbb{P}(\mathcal{E})$, with $\mathcal{E}$ normalized, then $e= - \deg \mathcal{E}$.

Moreover, since the base of the ruling is an elliptic curve $E$, then $e \in \{-1, 0, 1, 2, 3, 4, \ldots \}$.

We can exclude the case $e &gt;0$. Indeed, any divisor $D \in \textrm{Pic}(S)$ can be written as $a C_0 + b f$, where $f$ is the fibre of the ruling. Now the elliptic fibre of the morphism over $\mathbb{P}^1$ must satisfy  $(aC_0+bf)^2=0$, that is $b = \frac{1}{2}ae$. But if $e &gt;0$ then an effective divisor must also satisfy $b \geq ae$ (Hartshorne, Proposition 2.20 p. 382) and this is a contradiction.

Then the only possibility is $e \in \{-1, 0 \}$.

The case $e=-1$ corresponds to the second symmetric product $\textrm{Sym}^2(E)$; in this case $\mathcal{E}$ is the unique indecomposable rank two vector bundle on $E$ with $\deg \mathcal{E} =1$. Then the linear system $|4C_0-2f|$ is a pencil of elliptic curves. One can also write $S$ in the desired for (see Wil Sawin's first comment). 

The case $\mathcal{E}=0$ corresponds either to the trivial bundle (so $S$ is a product), or to $\mathcal{O}_E \oplus \mathcal{L}$, with $\deg \mathcal{L}=0$ or to $\mathbb{P}(\mathcal{E})$, where $\mathcal{E}$ is the unique indecomposible rank two vector bundle on $E$ such that $\deg \mathcal{E}=0$.

Let us consider first the case $\mathcal{O}_E \oplus \mathcal{L}$, with $\deg \mathcal{L}=0$. Then the curves in the elliptic pencil should correspond to elements of $h^0(aC_0)$, that is to sections of $\textrm{Sym}^a (\mathcal{O}_E \oplus \mathcal{L})$. Since we must have two independent sections, we obtain that  $\mathcal{L}$ is a torsion bundle. In this case the elliptic fibration do exist, and one can easily write $S$ in the desired form (see Will Sawin's second comment).

Finally, let us exclude the case $\mathbb{P}(\mathcal{E})$, with $\mathcal{E}$ indecomposable of degree $0$. Again, the curves in the elliptic pencil must correspond to elements of $h^0(aC_0)$, that is to sections of $\textrm{Sym}^a \mathcal{E}$. But from [Atiyah, Vector bundles on an elliptic curve, Theorem 9] one sees that $\textrm{Sym}^a  \mathcal{E}$ is again indecomposable of degree $0$, and that $h^0(\textrm{Sym}^a \mathcal{E})=1$. Thus one does not have two independent sections, and the elliptic pencil cannot exist.

Summing up, we obtain


  Let $p \colon S \to \mathbb{P}^1$ be an elliptic surface and assume that $S$ is ruled over an elliptic curve $E$. Then $S=\mathbb{P}(\mathcal{E})$, where either 
  
  $\bullet$ $\mathcal{E}$ is the unique indecomposable rank $2$ vector bundle of degree $1$ on $E$, or
  
  $\bullet$ $\mathcal{E}= \mathcal{O}_E \oplus \mathcal{L}$, where $\mathcal{L}$ is a (possibly trivial) torsion line bundle. 
  
  In both cases, we can also write $S$ in the desired form $S=(E \times \mathbb{P}^1)/G$. 


Surfaces ruled over elliptic curves
We know that Rakatan empire in Star Wars EU was one of the earliest hyperdrive capable species, and their hyperdrive was powered by the Dark Side of the Force:


  The first interstellar space drive known to make use of hyperspace was developed by the Rakata, who built their Infinite Empire around technology using the dark side of the Force to travel through hyperspace. (Wikia).


Has anyone used hyperdrive technology that was powered by the Force (any side) since the Rakata? 

It could have been a clone of Rakata's, or totally different, as long as it wasn't simply someone taking and using existing surviving Rakatan unit.

If not, was there an explicit in-universe explanation as to why the Force was never used to enhance Hyperspace travel? (the predictable "well the Rakata were screwed up via their use of Dark Side" doesn't seem to be plausible in a universe full of Sith, and assorted other Dark Force users).
