Took a bit of digging. You want to look at some older books, in this case Dirk J. Struik, Lectures on Classical Differential Geometry. A surface in $\mathbb R^3$ is indeed developable if and only if the Gauss curvature is identically zero. This is on page 91 of the Dover reprint. What the word developable means needs work: it means there is a one-parameter family of planes of which the surface is an envelope.
The concept of envelope is explicit in the relationship between Pascal's Theorem and Brianchon's Theorem in projective geometry: http://en.wikipedia.org/wiki/Pascal%27s_theorem and http://en.wikipedia.org/wiki/Brianchon%27s_theorem as a conic is either thought of as generated by a family (set) of points or by a family of tangent lines.
See Struik, table on page 72.
Alright, now, there are ruled surfaces such as the hyperboloid of revolution $x^2 + y^2 - z^2 = 1$ that are ruled. They are not developable, which is a stronger condition...in M. do Carmo, Differential Geometry of Curves and Surfaces, he goes so far as to say the a ruled surface with an extra condition is called developable, this is formula (9) on page 194... the ruled surface is given by
$$ \vec{x}(t,v) = \alpha(t) + v \, w(t) $$ which is formula (8). The condition to have a developable surface is that $(w, \dot{w}, \dot{\alpha})$ be always linearly dependent, which he writes as the determinant of the evident three by three matrix being identically zero. Note that this includes cones of revolution, with a singular point. It is not until page 408 that do Carmo proves that a complete surface with vanishing Gaussian curvature is a cylinder or a plane.
So, I feel that you are mixing two issues. Vanishing Gauss curvature shows ruled, but the result could have singularities. If complete, simply meaning no self intersections or singularities, the surface is a cylinder, well, or a plane. The hyperboloid is not developable.
A surface is a space whose points have each an open nghb homeomorphic to an open disc in $\mathbb R^2$. This is equivalent to have an open nghb homeomorphic to an open set in $\mathbb R^2$: the latter always contains an open disc! But the problem is that one cannot say $U=S\cap B$ is homeomorphic to an open disc, only that it contains some open set $W$ of $S$ homeomorphic to a disc. Then $W$ contains another $U'=S\cap B'$ and so on. Thus to argue with fundamental groups $\pi$ one must show that the inclusion $W\subset U$ induces an isomorphism $\pi(W^*)\to\pi(U^*)$ ($*$ means punctured, as usual). Then the first $\pi$ is that of a disc, $\mathbb Z$, and the second one that of the plane minus three points (see below). One needs something here, but Jordan is deeper than these remarks on fundamental groups (my opinion). In fact, to get a contradiction, it is enough to have a surjection $\pi(W^*)\to\pi(U^*)$ and knowing that the second group is not abelian.
Thus, we look at the inclusion $U'\subset U$. Let $\varepsilon$ and $\varepsilon'$ be the radii of $B$ and $B'$, and pick a sphere $E$ centered at $p$ of radius $\frac{1}{2}\varepsilon'$. In this situation the radial retraction of $\mathbb R^3\setminus\{p\}$ onto $E$ provides two deformations retracts $\rho:U^*\to S\cap E$ and $\rho|\!:U'^*\to S\cap E$. In fundamental groups we get isomorphisms
$$
\pi(U'^*)\to\pi(S\cap E)\quad\text{and}\quad \pi(U^*)\to\pi(S\cap E),
$$
which guarantee the homomorphism $\pi(U'^*)\to\pi(U^*)$ associated to the inclusion $U'^*\subset U^*$ (through which they factorize) is an isomorphism too. But this in turn factorizes through the inclusions $U'^*\subset W\subset U^*$ to give
$$
\pi(U'^*)\to\pi(W^*)\to\pi(U^*).
$$
Since this composition is bijective, the first arrow must be injective and second arrow onto. This latter fact is the minimum we needed, but remark that playing similarly with another disc $W'\subset U'$ we could obtain that $\pi(U'^*)\to\pi(W^*)$ is onto and then bijective.
Next we have the group $\pi(S\cap E)$, only to give some insight. The space $S\cap E$ is just a couple of circumferences meeting at two points. This is clearly homeomorphic to the union $F$ of an ellipse and a tangent inscribed circumference, sitting in a $\mathbb R^2$. Puncturing three points inside these curves in the obvious way we can deformation retract the plane onto $F$. Thus the homotopy model of $S\cap E$ is a $3$-punctured plane, and one must believe the fundamental group is $\mathbb Z*\mathbb Z*\mathbb Z$, not abelian.
Best Answer
Any (real analytic) developable surface must be (a subset of) a plane, a cylinder, a cone, or a tangent developable (the surface of tangent lines to a smooth curve in space — this surface will have a cuspidal edge along the curve). If by closed you mean a compact surface without boundary, there is none. Every developable surface must be ruled, and ruled surfaces are never compact.
See page 61 and Exercise 12 on page 42 of my differential geometry text.