The Faraday-Maxwell law says that
$$ \nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}$$
So, if the curl of the electric field is non-zero, then this implies a changing magnetic field. But if the magnetic field is changing then this "produces" (or rather must co-exist with) a changing electric field and is thus inconsistent with an electrostatic field.
The divergence of an electrostatic field can be zero (where there is no actual charge), but it cannot be zero everywhere. This because a conservative field (which is one that has a curl of zero) cannot form closed field lines; the field lines have to begin and end (on charges). Otherwise, you could move a test charge along a closed electric field line and return to the same point in space with the charge having gained kinetic energy because it was acted on by an electric force all the way around the loop in the direction of motion. But this is the antithesis of a conservative field.
I do not understand well the question. Are we discussing the existence of an electric field which is irrotational and solenoidal in the whole physical three-space or in a region of the physical three-space?
Outside a stationary charge density $\rho=\rho(\vec{x})$ non-vanishing only in a bounded region of the space, the produced static electric field is both irrotational and solenoidal. The specific form of $\rho$ is irrelevant.
If instead we are really considering the whole space without charges, a physically meaningful result is the following one.
The only possible static electric field (continously differentiable) defined in the whole three space without charges and such that its total energy is finite, i.e.,
$$\frac{\epsilon_0}{2}\int_{\mathbb{R}^3} \vec{E}(\vec{x})^2 d^3x <+\infty\tag{1}$$
is $\vec{E}(\vec{x})= \vec{0}$ everywhere.
Notice that $\vec{E}$ above is necessarily irrotational and solenoidal on the whole space, and vice versa, as a consequence of the Maxwell equations.
The assertion is true because $\nabla \times \vec{E}=0$ implies, in the whole space (simply connected), $\vec{E} = \nabla \phi$ for some scalar field $\phi$.
The requirement $\nabla \cdot \vec{E}=0$ implies $\Delta \phi=0$.
In turn, this condition implies that $\phi$ is smooth ($C^\infty$) and thus $\vec{E}$ admits second derivatives, in particular.
Obviously $\Delta E_k = \Delta \partial_k \phi = \partial_k \Delta \phi=0$.
Every component $E_k$ of $\vec{E}$ is therefore a harmonic function ($\Delta E_k =0$) on the whole space $\mathbb{R}^3$ and, due to (1), it also satisfies
$$\int_{\mathbb{R}^3} E_k(\vec{x})^2 d^3x <+\infty\quad k=1,2,3\:.$$
A known result of harmonic function theory states that a (real valued) harmonic function $f$ defined on the whole $\mathbb{R}^n$ such that the integral of $f^2$ is finite must be necessarily the zero function.
Best Answer
About divergence-free field lines
It isn't necessary for the lines of a divergence-free field to be closed. Zero divergence implies that the field lines can't end. Closed curves don't have endpoints, but there are other possibilities, such as lines that extend infinitely in both directions.
The relevant mathematical result here is Gauss's theorem. For any field, it says that the integral of its divergence in any region of space equals the integral of the field over the boundary of the region, the flux of field lines crossing the boundary surface. If you enclose in such a surface a point at which field lines start or end, the flux doesn't vanish, so the divergence of the field can't be zero.
For magnetic fields, the divergence is zero and then there aren't any ending points for its lines. For electric fields, the divergence equals the density of charge, and so its lines have ends at point with non-vanishing charge. The flux of lines across a closed surface is proportional to the charge enclosed by it.
A geometrical picture of irrotational fields
So we have a picture of electric field lines being born and dying at some (charged) points or regions of space. But we haven't used the property that it's irrotational.
If you're looking for some geometrical implication of zero rotational, a way of getting it is considering the surfaces that are orthogonal to field lines at every point. The existence of this surfaces, even locally, is not guaranteed if our field is not irrotational. However, they can be defined for some fields with non vanishing rotational. For example, for the magnetic field around an infinite straight wire, they would be the semiplanes limited by it. A surface defined in this way has an orientation: at each point one of its two orthogonal directions is determined by the direction of the field at that point. Although it's not a standard terminology, let me call them here "field surfaces". A nice analogy with field lines in the previous case will arise.
We can imagine field surfaces having boundaries; that is, there are curves in space at which these surface can end. For example, you can picture such a curve and a family of surfaces sharing it as a boundary, just as a family of field lines might end at a point. If we take an oriented closed curve wrapping around a boundary curve, the integral of the field over it will be proportional to the number of field surfaces crossing it, including a negative sign for each that has opposite orientation as the curve an a positive sign otherwise.
There's a generalization of Gauss's theorem that's useful in this case. It's called Stokes' theorem. The version we need now says that the integral over a surface of the rotational of a field equals the integral of the field over the boundary of the surface, a closed curve. Suppose that the surface over which we are integrating cuts a boundary curve. Therefore, by integrating the rotational we are counting the field surfaces that cut its boundary.
For electric fields (in the static case) the rotational is zero. Thus, following the reasoning above, we can conclude that electric field surfaces can't end.
A more precise formulation of the second part
The picture above can be formulated more precisely using differential forms and the Frobenius theorem. The idea is not to use the fact that a divergence-free field has a potential, but to work with geometrical objects which might be understood intuitively, as field lines are. In particular, we want to understand what theses field surfaces are and to learn something about their existence.
In general, any non-vanishing $1$-form $\omega$ on a $n$-dimensional manifold $M$ defines at each point $p\in M$ a $(n-1)$-dimensional subspace $S_p$ of the tangent space $T_pM=\mathbb{R}^3$ as $S_p=\{v\in T_pM:\;\omega(v)=0\}$. Such a smooth assignment of subspaces of the tangent space to every point is called a distribution.
One question we can ask about a distribution is whether we can find a family of $(n-1)$-dimensional submanifolds such that their tangent space at each $p$ is $S_p$. If we can, the distribution is said to be integrable. For our case, we will need a particular case of the Frobenius theorem: the statement that if a differential form is closed, then its associated distribution is integrable.
Now, the electric field can be seen as a $1$-form $E=E_xdx+E_ydy+E_zdz$ in the manifold $M=\mathbb{R}^3$. The distribution associated with $E$ is nothing more that the family of planes given by orthogonality to the electric field at a point. The condition that its rotational is zero is then just $dE=0$. So by the Frobenius theorem, the ($2$-dimensional) surfaces that are tangent to the planes $E$ defines exist. These are just the "field surfaces" that are used above.