My answer may not be as complete as desired -- so far I've just done the case of the candle lying somewhere above the cylinder, at a height $d$ above the top of the cylinder, with horizontal radial distance $s$ from the center.
Let $h$ be the height of the cylinder, and $r \ge s$ be the radius of the cylinder. Parametrize the circular top of the cylinder with $\theta \in [0,2\pi)$ . Note that the segment connecting the candle's projection onto the cylinder, with a point $\theta$ on the circle, has length $\sqrt{r^2+s^2-2rs\cos\theta}$. Let the length of the shadow on the ground, considered radially from the candle, be $R(\theta)$. Then seeing that the triangles formed by the light ray extending from the candle to the edge of the cylinder's top, and that to the ground, are similar, comparing tangents we find that $ R(\theta) = \left(1+\frac{h}{d}\right)\sqrt{r^2+s^2-2rs\cos\theta} ~~. $
To calculate the area of this shadow, we note that the area of a differential sector is $\frac{1}{2}R^2d\theta$ , and compute $ A_s = \int_0^{2\pi} \frac{1}{2}\left(1+\frac{h}{d}\right)^2(r^2+s^2-2rs\cos\theta)d\theta = \pi(s^2+r^2)\left(1+\frac{h}{d}\right)^2 ~~. $ Finally, removing the cylinder yields the desired area, $A_s - \pi r^2$.
When $s=0$, we retrieve the correct candle-above-the-center case, $\pi r^2 \left(1+\frac{h}{d}\right)^2 - \pi r^2$.