Here is another way to look at it that you might find useful:  Essentially, you are asking the following problem:  Given a cylinder $C = (0,R)\times S^1$ and a positive $2$-form $\omega = f(r,\theta)\,\mathrm{d}r\wedge\mathrm{d}\theta$, one wants to describe the diffeomorphisms of $C$ that preserve the $2$-form $\omega$ and the foliation given by the level curves of $r$.  For simplicity, I will assume that one only cares about the symplectomorphisms of $(C,\omega)$ that fix the two 'ends' and that the total $\omega$-area of $C$ is finite; I'll say what to do about the larger group and the infinite area case at the end. 

First, make a change of variables:  Let 
$$
\rho = \frac1{2\pi}\int_0^r\int_0^{2\pi} f(\tau,\theta)\ \mathrm{d}\tau\wedge\mathrm{d}\theta
$$
Then $\rho:(0,R)\to (0,P)$ is a diffeomorphism, where $2\pi P\le\infty$ is the total $\omega$-area of $C$, and we can regard $\rho$ as a function on $C$ with the same level sets as $r$.  Obviously, any symplectomorphism of $(C,\omega)$ that preserves the given foliation (and the ends of $C$) will preserve $\rho$, so one might as well replace $r$ by $\rho$ (and set $R=P$).  Thus, let's just assume that $\rho=r$.  

One now has that
$$
\int_0^r\int_0^{2\pi} f(\tau,\theta)\ \mathrm{d}\tau\wedge\mathrm{d}\theta
= \int_0^r\int_0^{2\pi} \mathrm{d}\tau\wedge\mathrm{d}\theta,
$$
and, differentiating with respect to $r$, this gives
$$
\int_0^{2\pi} \bigl(f(r,\theta)-1\bigr)\ \mathrm{d}\theta  = 0
$$
for all $r$.  In particular, it follows that, setting
$$
\phi(r,\theta) = \int_0^\theta \bigl(f(r,\psi)-1\bigr)\,\mathrm{d}\psi,
$$
one has $\phi(r,\theta+2\pi) = \phi(r,\theta)$, so that $\phi$ is well-defined on $C$.  Moreover,
$$
\mathrm{d}r\wedge\mathrm{d}\phi = \bigl(f(r,\theta)-1\bigr)\,\mathrm{d}r\wedge\mathrm{d}\theta,
$$
so $\mathrm{d}r\wedge\mathrm{d}(\theta+\phi) = f(r,\theta)\,\mathrm{d}r\wedge\mathrm{d}\theta = \omega$.  Setting $\psi = 
\theta+\phi$, one now has $\omega = \mathrm{d}r\wedge\mathrm{d}\psi$.

In the coordinates $(r,\psi)$ on $C$, it is now easy to describe the symplectomorphism of $(C,\omega)$ that preserve the foliation defined by $\mathrm{d}r=0$, they are just the diffeomorphisms of the form
$$
F(r,\psi) = \bigl(r,\psi+g(r)\bigr),
$$
where $g$ is an arbitrary (differentiable) function of $r$.  Strictly speaking, these are the ones that preserve the ends of $C$.  To get the ones that exchange the ends of $C$ while preserving the given foliation, one can compose an $F$ of the above form with the involution
$$
H(r,\psi) = (R-r,-\psi).
$$

Now, in the case that the $\omega$-area of $C$ is infinite, one has to slightly modify the construction by integrating from a fixed $r=r_0\in(0,R)$ instead of from an 'end'.  I.e., one defines
$$
\rho = \frac1{2\pi}\int_{r_0}^r\int_0^{2\pi} f(\tau,\theta)\ \mathrm{d}\tau\wedge\mathrm{d}\theta.
$$
Then $\rho:(0,R)\to\mathbb{R}$ is a diffeomorphism onto its image, which might be all of $\mathbb{R}$.  In fact, if the image is not all of $\mathbb{R}$, then at least one of the two 'ends' has finite area and one can reduce to the case that the image is either $(0,\infty)$ or $(-\infty,0)$ and the arguments go through essentially unchanged.  When the image is all of $\mathbb{R}$, one can still replace $r$ by $\rho$ as a coordinate on $C$ and the one still has
$$
\int_0^{2\pi} \bigl(f(r,\theta)-1\bigr)\ \mathrm{d}\theta  = 0,
$$
and again, everything goes through as before, but now, when one gets to the coordinate system $(r,\psi)$ in which $\omega = \mathrm{d}r\wedge\mathrm{d}\psi$, the group of diffeomorphisms that preserve the level sets of $r$ and the $2$-form $\omega$ are now generated by
$$
F(r,\psi) = \bigl(r+c,\psi+g(r)\bigr)
$$
where $c$ is any constant and $g$ is any (smooth) function of $r$ and 
$$
H(r,\psi) = (-r,-\psi).
$$