Here are a lot of things mixed up in this thread! So let's clarify a few of them.
First of all an electric field can have a curl, if there is a time-dependent magnetic field present, because due to Faraday's Law, which is one of the fundamental Maxwell equations of the electromagnetic field:
\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}.
In static or stationary situations, i.e., when only time-independent fields are present, thus the curl vanishes.
Now you have given vector fields given in terms of cylindrical coordinates, (s,\varphi,z). The naming s is uncommon, but let me keep your notation. Let's first clarify the notation a bit. The components in cylinder coordinates are given in terms of the coordinate's basis vectors, i.e., your two fields are
1. \vec{E}_1=s \vec{e}_{\varphi},
2. \vec{E}_2=(1/s) \vec{e}_{\varphi}.
Next you have to think about the curl, which is a vector operator, and you have to express it in terms of the coordinates used. For the cylinder coordinates, it looks as follows:
<br />
\begin{split}<br />
\vec{\nabla} \times \vec{E} =& \vec{e}_s \left (\frac{1}{r} \frac{\partial E_z}{\partial \varphi} - \frac{\partial E_{\varphi}}{\partial z} \right ) \\<br />
&+ \vec{e}_{\varphi} \left (\frac{\partial E_s}{\partial z}-\frac{\partial E_z}{ \partial s} \right ) \\<br />
&+ \vec{e}_z \frac{1}{s} \left (\frac{\partial(s E_{\varphi})}{\partial s}-\frac{\partial E_s}{\partial \varphi} \right ).<br />
\end{split}<br />
Now, luckily your vectors are pretty simple, and you can easily plug them in and take the derivatives. You finally get
\vec{\nabla} \times \vec{E}_1=2 \vec{e}_z, \quad \vec{\nabla} \times \vec{E}_2=0.
This shows you that \vec{E}_1 cannot be an electrostatic field, because it's curl is not vanishing. It also cannot be a gradient field.
For \vec{E}_2 it's more tricky, because apparently is looks as if it is a gradient field, because its curl vanishes. In fact it is locally a gradient field, namely in any simply connected neighborhood of a regular point of the field, but as a whole the region, where this field is defined is not a connected region in \mathbb{R}^3, because obviously there is trouble for s=0, which is the z axis.
Now you must become even more alarmed, because cylindrical coordinates are not covering the entire \mathbb{R}^3! The coordinates are singular precisely for s=0, because there the Jacobi determinant, which is just s between the cylinder and Cartesian coordinates vanishes. Thus, we have to check, whether the apparent singularity is only due to these trivial coordinate singularities or whether there are true singularities.
In this case, the most simple way to check this is to go back to Cartesian coordinates. To that end we just need
s=\sqrt{x^2+y^2}, \quad \vec{e}_{\varphi}=-\vec{e}_x \frac{y}{s} + \vec{e}_y \frac{x}{s}.
So there is indeed a real singularity along the z axis. The \mathbb{R}^3 with the z axis taken out obviously is not simply connectec, because any closed curve that encircles the z axis cannot be contracted in any continuous way to a single point, because you somehow always have to cross the z axis, which is fobidden, because it's taken out of the region, where the field is defined.
You can also check this in another way. Just calculate a line integral along a closed line around the z axis. It's simple to prove with help of Stokes's integral theorem that the result only depends on the number of windings around the z axis, because the curl of the field vanishes everywhere except along the z axis, where the field is singular and undefined. So we can choose any curve we like, and here it's most easy to choose a circle parallel in xy plane with the origin of the coordinate system as its center. We can again switch to cylinder coordinates for that calculation, because we are working off the [itez]z[/itex] axis. This gives for a counter-clockwise oriented circle
\int_{C_r} \mathrm{d} \vec{s} \cdot \vec{E}_2=\int_0^{2 \pi} \mathrm{d} \varphi s \frac{1}{s}=2 \pi.
If we had wound the circle n times around the z axis, we'd had gotten 2 \pi n.
Finally we can ask, how to realize such a field as a electrostatic field, because except along the z axis its curl vanishes, and thus there is may describe an electrostatic field!
To that end let's find the local potential, i.e., a scalar field \Phi such that
\vec{E}_2=-\vec{\nabla} \Phi.
The gradient in cylinder coordinates reads
\vec{\nabla} \Phi=\vec{e}_s \frac{\partial \Phi}{\partial s} + \vec{e}_{\varphi} \frac{1}{s} \frac{\partial \Phi}{\partial \varphi} + \vec{e}_{z} \frac{\partial \Phi}{\partial z}.
Thus you get for the potential of \vec{E}_2
\frac{\partial \Phi}{\partial s}=\frac{\partial \Phi}{\partial z}=0, \quad \frac{1}{s} \frac{\partial \Phi}{\partial \varphi}=-\frac{1}{s}.
From the first equation you see that \Phi can depend only on \varphi, and the last equation tells you that
\Phi=-\varphi.
Now, this clearly shows, what's going on! The potential is not simple valued. You can define \varphi to take values in any interval of length 2 \pi to cover the whole \mathbb{R}^3 with the z axis taken out. Let's choose the interval (-\pi,\pi). This covers the whole \mathbb{R}^3 with the half-plane x<0, \quad y=0 taken out, and on the "upper" boundary the potential takes the value -\pi and on the "lower" boundary it takes the value +\pi.
Thus you have a voltage difference along this half-plane, leading to this curly electrostatic field. Of course, this is a highly academic interpretation, because of course you cannot realize this configuration in practice, because you cannot set up an infinitely large half-plane with two metal foils separated by an infinitesimally thin insulator between them and hooking it up on a battery at infinity.
