To make things easier, first change the $x,y,z$ coordinates to a new system of Cartesian coordinates $x',y',z'$ parallel to the original one but with the origin at $(0,0,-1)$, so $x'=x$, $y'=y$, $z'=z+1$. In the new coordinate system the vector $\vec g$ is:
$$\vec g = \frac{(x',y',z')}{(x^{\prime 2}+y^{\prime 2}+z^{\prime 2})^{3/2}} = \frac{\vec r}{r^3}\,,$$
where $\vec r=(x',y',z')$ is the radial vector field (from the new origin of coordinates) and $r$ its magnitude. This brings up the symmetry of the problem (and its physical connection; $\vec g$ is proportional to the electric field of a static electric charge at the new origin of coordinates, or to the the magnetic field produced by a Dirac's magnetic monopole at that point). Although you can work in Cartesian coordinates, to take advantage of the symmetry it is probably easier to work using standard spherical coordinates $r,\theta,\phi$, and that is what I am going to do, later converting back to Cartesian coordinates. Before we start it is a good idea to check that $\nabla\cdot\vec g=0$, otherwise the problem has no solution. But the divergence of this vector field is well known ($\vec g$ is essentially the gradient of the Green's function of the Poisson equation):
$$ \nabla\cdot\frac{\vec r}{r^3} = 4\pi\delta^3(\vec r)\,.$$
Although outside the origin the divergence is zero, because it is not zero everywhere in $R^3$, it is not possible to find a vector $\vec f$ that is non-singular everywhere $R^3$ minus the origin and such that $\nabla\times\vec f=\vec g$. However it is going to be possible to find $\vec f$ for the required region $z>0$. Note that the solution to this equation is far from unique, because adding to a solution any gradient also gives a solution. We need only to find one solution, all other solution are obtained form this one by adding an arbitrary gradient. To make the calculations easier we can impose some reasonable condition that will restrict the number of available solutions (physicsits call this to "fix the gauge"). For example, if you want to work in Cartesian coordinates you could require your solution to have $f_z=0$ (physicists call this choice the "axial gauge"). Because $\vec g$ is radial, and the radial component of $\vec f$ does not contribute to it, I am going to make the radial component of $\vec f$, $f_r$ equal to zero. With this choice, using the expression of the curl in spherical coordinates we have for $f_\theta$ and $f_\phi$:
$$
\begin{align}
\frac{1}{r\sin{\theta}}\left[\frac{\partial}{\partial\theta}\left(\sin\theta f_\phi\right)-\frac{\partial f_\theta}{\partial\phi}\right] &= \frac{1}{r^2} \\
\frac{\partial}{\partial r}\left(rf_\phi\right) &= 0 \\
\frac{\partial}{\partial r}\left(rf_\theta\right) &= 0
\end{align}
$$
The last two equations show that in this gauge ($f_r=0$), $f_\theta$ and $f_\phi$ have to be proportional to $1/r$, which makes also the radial part of the first equation correct. In fact, with this in mind, it is easily seen by inspection of the first equation that the following choice is a solution:
$$f_\phi = \frac{C-\cos\theta}{r\sin\theta},~~f_\theta = 0\,,$$
where $C$ is some constant. Because we want the solution to be valid if $z>0$ we
must choose $C=1$. In other words, if $\hat u_\phi$ is the unit vector along the $\phi$ direction, then:
$$\vec f = \frac{1-\cos\theta}{r\sin\theta}\hat u_\phi$$
is a solution to the problem in spherical coordinates. The only thing that remains is to convert back to Cartesian coordinates. If $\hat u_x, \hat u_y, \hat u_z$ are the unit vectors along the Cartesian axes (old and new!), and using:
$$\hat u_\phi = -\sin\phi\,\hat u_x+\cos\phi\,\hat u_y$$
and
$$\cos\phi = \frac{x'}{r\sin\theta},~~\sin\phi = \frac{y'}{r\sin\theta},
~~\cos\theta = \frac{z'}{r},~~r^2\sin^2\theta = x'^2+y'^2$$
we get:
$$\vec f = \frac{y'(z'-r)}{r(x'^2+y'^2) }\hat u_x -\frac{x'(z'-r)}{r(x'^2+y'^2) }\hat u_y\,.$$
Reverting to the original $x,y,z$ coordinate system (just replace $x'~y'$ by $x~y$, and $z'$ by $z+1$) we get the final answer:
$$\vec f = \frac{y(z+1-r)\,\hat u_x-x(z+1-r)\,\hat u_y}{r(x^2+y^2)}$$
with
$$ r=\sqrt{x^2+y^2+(z+1)^2}\,.$$
Note that this solution satisfy the physicists "axial gauge" $f_z=0$, although I was not looking for it. Also note that this solution is regular for $z>-1$, as required, and becomes singular in the half line $x=y=0, z<=-1$.
Best Answer
If
$\nabla \times \vec E = 0, \tag 1$
we know from a standard result that
$\vec E = \nabla \phi \tag 2$
for some scalar function $\phi$.
If the divergence of $\vec E$ is also a known function $\rho$,
$\nabla \cdot \vec E = \rho, \tag 3$
then combining (2) and (3) we obtain
$\nabla^2 \phi = \nabla \cdot \nabla \phi = \nabla \cdot \vec E = \rho, \tag 4$
which can in principle be solved for $\phi$ (assuming we admit appropiate boundary conditions on $\phi$; $\phi(x) \to 0$ sufficiently fast as $\vert x \vert \to \infty$ is often used; Jackson's book covers this, mos' likely); thus, we may discover $\vec E$ from (2).
We observe that such a solution is not unique; indeed, let $\psi$ be any harmonic function,
$\nabla \cdot \nabla \psi = \nabla^2 \psi = 0; \tag 5$
then
$\nabla^2(\phi + \psi) = \nabla^2 \phi + \nabla^2 \psi = \rho + 0 = \rho, \tag 6$
and if
$\vec E = \nabla (\phi + \psi), \tag 7$
we still obtain
$\nabla \times \vec E = \nabla \times \nabla (\phi + \psi) = \nabla \times \nabla \phi + \nabla \times \nabla \times \psi = 0, \tag 8$
since the curl of any gradient vanishes. Also,
$\nabla \cdot \vec E = \nabla \cdot (\nabla \phi + \nabla \psi) = \nabla^2 \phi + \nabla ^2 \psi = \nabla^2 \phi = \rho; \tag 9$
these last two equations show that we may transform any solution according to
$\phi \to \phi + \psi, \tag 9$
$\vec E = \nabla \phi \to \nabla \phi + \nabla \psi, \tag{10}$
$\psi$ as in (5), and preserve the divergence and curl of $\vec E$; so any solution is not unique; uniqueness may be attained by specifying appropriate boundary conditions on $\phi$ and $\psi$ which can then become unambiguously determined.
The above discussion addresses the relatively simple case (1), (3); we can, in fact, also address the significant generalization
$\nabla \times \vec E = \vec F, \; \nabla \cdot \vec E = \rho, \tag{11}$
where $\vec F$ is a pre-specified vector field; the situation is more complicated since we may no longer assume $\vec E$ is a gradient as in (1)-(2). In this case we instead invoke the vector calculus identity
$\nabla \times (\nabla \times \vec A) = \nabla (\nabla \cdot \vec A) - \nabla^2 \vec A, \tag{12}$
where the Laplacian operator $\nabla^2$ occurring on the right-hand side is understood to act component-wise on $\vec A$; thus we have
$\nabla \times \vec E = \vec F \tag{13}$
leading to
$\nabla \times (\nabla \times \vec E) = \nabla \times \vec F, \tag{14}$
which we trasform according to (12):
$\nabla (\nabla \cdot \vec E) - \nabla^2(\vec E) = \nabla \times \vec F; \tag{15}$
by virtue of (3) we write this as
$\nabla \rho - \nabla^2 \vec E = \nabla \times \vec F, \tag{16}$
whence we find
$\nabla^2 \vec E = \nabla \rho - \nabla \times \vec F, \tag{17}$
which we may in principle solve component-wise for $\vec E$; for example,
$\nabla^2 E_x = \dfrac{\partial \rho}{\partial x} - \left ( \dfrac{\partial F_z}{\partial y} - \dfrac{\partial F_y}{\partial z} \right) = \dfrac{\partial \rho}{\partial x} - \nabla^2 E_x = \dfrac{\partial \rho}{\partial x} - \dfrac{\partial F_z}{\partial y} + \dfrac{\partial F_y}{\partial z}; \tag{18}$
of course, we need boundary conditions and perhaps certain restrictions on $\rho$ and $\vec F$; but I'll leave it to folks like David Jackson to explain such matters.
Apparently solutions to (11), (18) are still not unique; it is still true that adjusting $\vec E$ by the gradient of a harmonic function $\psi$ gives rise to another solution; of course, we may determine such $\psi$ uniquely via appropriate boundary conditions, and then a certain uniqueness is attained for $\vec E$.