Here are two relevant counterexamples:
Let $R$ be a ring and $I$ an ideal, then the morphism $\operatorname{Spec} R[x]/(x-x^2,Ix) \to \operatorname{Spec} R$ has connected fibers outside $V(I)$ but disconnected fibers inside $V(I)$. This suggests that some kind of flatness condition is unavoidable.
Let $R$ be a non-Henselian local ring and let $f$ be an irreducible polynomial that factors into coprime polynomials modulo the maximal ideal. Then $\operatorname{Spec} R[x]/f(x) \to \operatorname{Spec} R$ has connected generic fiber but disconnected special fiber. This suggests that some sort of geometric connectedness condition fro the generic fiber is unavoidable.
But we do not always need the generic fiber to be "completely" geometrically connected. Let $Y=\operatorname{Spec} k[[x]]$, then if the generic fiber of a flat proper morphism $f: X \to Y$ is connected, then the special fiber is connected. This is because $f_* \mathcal O_X$ is a finite flat algebra. After inverting $x$, and ignoring nilpotents, it injects into a finite field extension of $k((x))$, which must be the field of fractions of a complete local ring, so it itself is a complete local ring.
Presumably we should formulate the most general version locally, with $Y$ the spectrum of a local ring $R$, for simplicity. A sufficient flatness condition is that the morphism is flat, or that $f_* \mathcal O_X$ is flat. A sufficient connectedness condition is that the fiber over the "Henselian generic point", the spectrum of the field of fractions of the Henselization or completion of $R$, is connected.
We can see this by reducing the connectedness of $X$ to the connectedness of the second half of Stein factorization via Zariski's connectedness theorem. We reduce to the case where $R$ itself is Henselian by completing. Then $f_* \mathcal O_X$ is an $R$-algebra which injects into a connected algebra over the field of fractions of $R$, so modulo nilpotents, which we can safely ignore, it injects into a finite field extension of $R$.
Assume that the special fiber is disconnected, that is, $f_* \mathcal O_X/m$ contains an idempotent mod $m$. Lift that idempotent to $f_* \mathcal O_X$, and compute its minimal polynomial. This is irreducible, so by Hensel's Lemma it has a unique root mod $m$, which must be either $0$ or $1$ since $x^2-x$ divides the minimal polynomial mod $m$, so the idempotent is equal to either $0$ or $1$, a contradiction.
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 $e >0$. 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 $e >0$ 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 $S=(E \times \mathbb{P}^1)/G$.
Best Answer
I thought I'd expand my earlier comment, which was not all that clear, and I'm not even sure where you'd look it up. Let's say that a ruled surface over a smooth curve $C$ is smooth projective morphism $f:X\to C$ all of whose fibres are isomorphic to $\mathbb{P}^1$. Then one checks that $\omega_{X/C}^{-1}$ is relatively very ample. So we get an embedding $X\to \mathbb{P}(f_*\omega_{X/C}^{-1})$ over $C$ as a family of conics. Assuming that you are over an algebraically closed field, you can apply Tsen to see that the generic fibre has a rational point. This implies that there is a rational section $C\dashrightarrow X$. Since $C$ is a curve, this extends to an honest section.
An alternative argument is to note that the obstruction to $X=\mathbb{P}(E)$, for some vector bundle $E$, lies in the Brauer group of $C$ via $$H^1(C_{et},GL_2)\to H^1(C_{et},PGL_2)\to H^2(C_{et},G_m)$$ and this vanishes by Tsen. One gets a section using this as well, and much more.