As you noted, that's not the Schwarzschild metric in cylindrical coordinates.
In spherical coordinates, where the corresponding Cartesian coordinates would be
$$(x,y,z) = r (\sin θ \cos φ, \sin θ \sin φ, \cos θ),$$
the metric is given by the line element
$$\frac{r}{r - r_s} dr^2 + (r dθ)^2 + (r \sin θ dφ)^2 - \frac{r - r_s}{r} (c dt)^2.$$
To express the metric in cylindrical coordinates, where
$$(x,y) = ρ (\cos φ, \sin φ),$$
set
$$(ρ,z) = r (\sin θ, \cos θ).$$
Then
$$ρ dρ + z dz = r dr, \hspace 1em z dρ - ρ dz = r^2 dθ, \hspace 1em ρ dφ = r \sin θ dφ, \hspace 1em r = \sqrt{ρ^2 + z^2},$$
and the line element becomes:
$$\frac{\sqrt{ρ^2 + z^2}}{\sqrt{ρ^2 + z^2} - r_s} \frac{(ρ dρ + z dz)^2}{ρ^2 + z^2} + \frac{(z dρ - ρ dz)^2}{ρ^2 + z^2} + (ρ dφ)^2 - \frac{\sqrt{ρ^2 + z^2} - r_s}{\sqrt{ρ^2 + z^2}} (c dt)^2.$$
There is also the underlying question of what the cylindrical *generalization* of the Schwarzschild metric is. That's the Kerr metric which, when given in Boyer-Lindquist coordinates, is:
$$ds^2 = Σ \left(\frac{dr^2}{Δ} + dθ^2\right) + \left(r^2 + a^2\right) (\sin θ dφ)^2 + \frac{r_s r}Σ \left(c dt - a \sin^2 θ dφ\right)^2 - (c dt)^2,$$
where the corresponding Cartesian coordinates are
$$(x,y,z) = \left(\sqrt{r^2 + a^2} \sin θ \cos φ, \sqrt{r^2 + a^2} \sin θ \sin φ, r \cos θ\right),$$
and
$$r_s = \frac{2GM}{c^2}, \hspace 1em a = \frac{J}{Mc}, \hspace 1em Σ = r^2 + (a \cos θ)^2, \hspace 1em Δ = r^2 - r_s r + a^2,$$
and this
$r_s$ being the same as the
$r_s$ for the Schwarzschild metric.
This describes a source with angular momentum
$J$, mass
$M$, in relativity, with
$c$ being the in-vacuo light speed. You can work that what that comes out to in cylindrical coordinates, with the coordinates modified to the following form:
$$ρ = \sqrt{r^2 + a^2} \sin θ, \hspace 1em z = r \cos θ,$$
and use the cylindrical version of the Schwarzschild metric to check this against the
$J = 0$ (and
$a = 0$) case.
What your are encountering here is a core feature of general relativity: time is inherently, and fundamentally a local quantity. While each observer will have an unambiguously defined local proper time, there is generally no unique way to extend this notion of time to the rest of the universe. The global time coordinates $t$ and $t'$ both reduce to the proper time of a distant stationary observer, but differ away from that observer. Both can equally claim to be a generalization of the time of the observer, and so can an infinity of other generalizations.
Consequently, there is no (unambiguous) notion in general relativity of "the time at some event according to some (distant) observer".
The best one can hope to do, is describe what a particular observer actually observes. This generally requires considering how this information got to the observer. Typically this is done by transmitting light rays of some type, and therefore requires tracing how a light ray travels from the event to the observer. Generally, this is not as easy as simply comparing time coordinates at the different points, although some coordinates systems are better adapted to such an analysis as others.
Neither Schwarzschild nor advanced Eddington-Finkelstein coordinates (as used in the OP) are particularly well adapted to analyzing the signal received by a distant observer coming from an object crossing the (future) event horizon of a Schwarschild black hole. For this it is better to use retarded Eddington-Finkelstein coordinates. In these coordinates one straigthforwardly sees that it takes an infinite amount of time for a distant observer to receive the last signal sent by an object as it crosses the horizon, i.e. the distant observer will never (according to his own time) see the object cross the horizon.
Best Answer
The transformation rule can be found here and here:
$${\rm dt = d\tau+dr \ v} / \hat g_{\rm tt} \ , \ \ {\rm v = -c \ \sqrt{r_s/r}} \ , \ \ 1/\gamma = \rm \sqrt{1-v^2/c^2}$$
with that you transform the old coordiantes $\rm \hat x$ to the new ones $\rm \bar x$:
$$\bar g_{\mu \nu} = \sum_{\sigma, \kappa} \ \hat g_{\sigma \kappa} \ \rm \frac{\partial \hat x^{\sigma}}{\partial \bar x^{\mu}} \ \frac{\partial \hat x^{\kappa}}{\partial \bar x^{\nu}}$$
so the metric in regular Schwarzschild/Droste coordinates (with time $\rm \hat x^0 = t$) where the local observers are stationary with respect to the black hole:
$$\hat g_{\mu \nu} = \left( \begin{array}{cccc} \rm c^2/\gamma^2 & 0 & 0 & 0 \\ 0 & \rm -\gamma^2 & 0 & 0 \\ 0 & 0 & \rm -r^2 & 0 \\ 0 & 0 & 0 & \rm -r^2 \sin ^2 \theta \\ \end{array} \right)$$
transforms to Gullstrand/Painlevé coordinates (with time $\rm \bar x^0 = \tau$) where the local observers are free falling raindrops with the negative escape velocity $\rm v$:
$$\bar g_{\mu \nu} = \left( \begin{array}{cccc} \rm c^2/\gamma^2 & \rm v & 0 & 0 \\ \rm v & \rm -1 & 0 & 0 \\ 0 & 0 & \rm -r^2 & 0 \\ 0 & 0 & 0 & \rm -r^2 \sin ^2 \theta \\ \end{array} \right)$$
where the covariant spatial components are euclidean:
$$\bar g_{\rm i j} = \left( \begin{array}{cccc} \rm -1 & 0 & 0 \\ 0 & \rm -r^2 & 0 \\ 0 & 0 & \rm -r^2 \sin ^2 \theta \\ \end{array} \right)$$
which is just flat space in spherical $\{ \rm r, \ \theta, \ \phi \}$ coordinates:
Here are two such free falling raindrops in Gullstrand/Painlevé coordinates sending signals to each other when they are half way through the black hole (at $\rm r=r_s/2=1$, photons in green):
If you take ${\rm dt = du}\pm{\rm dr \ c} / \hat g_{\rm tt}$ or alternatively ${\rm dt = d{}_T\pm dr \ v^2/c} / \hat g_{\rm tt}$ instead (with $\rm {}_T=u \mp r$) you get the Schwarzschild metric in ingoing or outgoing Eddington/Finkelstein coordinates where radial photons have a constant coordinate velocity of $\rm dr/d{}_T=-c$, while in Gullstrand/Painlevé coordinates the free fallers have a coordinate velocity of $\rm dr/d\tau=v$.
In regular Schwarzschild/Droste coordinates on the other hand, at $\rm r=r_s$ you get $\rm dr/dt=0$ for all particles and photons since the bookkeeper's coordinate time $\rm t$ freezes at the horizon.