curl of what yields $(0,s^{-1},0)$ in cylindrical coordinates? In cylindrical coordinates $(s,\theta,z)$, what function $\mathbf{A}$ has the property $$\nabla\times \mathbf{A} = (0, \frac{1}{s} , 0) $$
I know generally that $$\nabla\times \mathbf{A} = \left(\frac{1}{s}\frac{\partial A_3}{\partial \theta}-\frac{\partial A_2}{\partial z}, \frac{\partial A_1}{\partial z}-\frac{\partial A_3}{\partial s} , \frac{1}{s} \left( \frac{\partial}{\partial z}(s\cdot A_2) - \frac{\partial A_1}{\partial \theta}\right)\right) $$
Is there some better way of solving for $\mathbf{A}$ other than a lot of ugly equations?

There is a path integral formula to construct the inverse of the exterior derivative operator for closed differential forms, for curl in three dimensions:
$$
\newcommand{\A}{\mathbf{A}}\newcommand{\ps}{\boldsymbol{\psi}} \newcommand{\p}{\mathbf{p}} \mathscr{R}(\ps) = -(\p - \p_0)\times \int^1_0 \ps\big(\p_0 + t(\p- \p_0)\big)t\,dt,\tag{1}
$$
where $\p = (x,y,z)$ and $\ps$ has to be divergence free. We have that $\mathscr{R}(\ps) $ satisfies:
$$
\nabla \times \mathscr{R}(\ps) = \ps.
$$

The vector field you gave, $(0,1/s,0) = (0,\dfrac{1}{\sqrt{x^2+y^2}},0)$, is not divergence free, there does not exist any vector potential $\A$ satisfying:
$$
\nabla \times \A = (0,\frac{1}{s},0). 
$$
However, we can just modify a little bit to make your question a well-posed problem, we can move $1/s$ to the $z$-component, and it is divergence free:
$$
\nabla \cdot \left(0,0,\frac{1}{\sqrt{x^2+y^2}}\right) = 0,
$$
and 
$$
\nabla \times \A = (0,0,\frac{1}{s})
$$
has a solution (not unique though).

Let $\ps$ be the above vector, and in (1) we take $\p_0$ as the origin we have:
\begin{align}
\A = &\mathscr{R}(\ps) = -\p\times \int^1_0 \ps(t\p)t\,dt
\\
&= -(x,y,z)\times \left(0,0,\frac{1}{\sqrt{x^2+y^2}}\right)
\\
&=  \left(-\frac{y}{\sqrt{x^2+y^2}},\frac{x}{\sqrt{x^2+y^2}},0\right).\tag{2}
\end{align}
We can check:
\begin{align}
\nabla \times\A &= \left(0,0, \frac{\partial }{\partial x}\Big(\frac{x}{\sqrt{x^2+y^2}}\Big) + \frac{\partial }{\partial y}\Big(\frac{y}{\sqrt{x^2+y^2}}\Big)\right)
\\
&= \left(0,0, \frac{\sqrt{x^2+y^2} - x\frac{x}{\sqrt{x^2+y^2}}}{x^2+y^2} + \frac{\sqrt{x^2+y^2} - y\frac{y}{\sqrt{x^2+y^2}}}{x^2+y^2}\right)
\\
& = \left(0,0,\frac{1}{\sqrt{x^2+y^2}}\right).
\end{align}
Or you can rewrite (2) in cylindrical coordinates:
$$
\A = (-\sin\theta, \cos\theta,0),
$$
and use the curl formula you gave.
