Application of Green's Theorem when undefined at origin

By the definition of open set, since $(0,0) \in \Omega$, there is an $r>0$ such that the disk of radius $r$ centred at $(0,0)$, $B_r(0,0)$ is contained in $\Omega$. Hence you can define $\Omega_r = \Omega \setminus B_r(0,0)$, which has two distinct boundaries, $\partial \Omega_r$ and $\partial B_r(0,0)$.

Now, as you say, we can use Green's Theorem on $\Omega_r$, so $$ 0 = \iint_{\Omega_r} (\partial_x Q - \partial_y P)\, dx\,dy = \int_{\partial\Omega_r} P \, dx + Q \, dy = \left( \int_{\partial\Omega} -\int_{x^2+y^2=r^2} \right) (P \, dx + Q \, dy), $$ because the boundary of $B_r(0,0)$ is traversed in the opposite direction because we have to keep the interior of $\Omega_r$ on the left.

Therefore we have to compute $\int_{x^2+y^2=r^2} (P \, dx + Q \, dy)$. Since this is just a circle, we parametrise it by $x=-r\cos{t}$, $y=r\sin{t}$, $dx=r\cos{t} \, dt$, $ dy=r\sin{t} \, dt $: $$\int_{x^2+y^2=r^2} (P \, dx + Q \, dy) = \int_0^{2\pi} \frac{-r\sin{t}}{r^2(\cos^2{t}+\sin^2{t})} (-r\sin{t}) \, dt + \frac{r\cos{t}}{r^2(\cos^2{t}+\sin^2{t})} (r\cos{t}) \, dt \\ = \int_0^{2\pi} 1 \, dt = 2\pi. $$

If your boundary is smooth, then you may parametrize it by $x(t) = r(t) \cos t$ and $y(t) = r(t) \sin t$, with $t \in [0,2\pi]$. Then your differential form $P \Bbb d x + Q \Bbb d y$ becomes

$$\frac {-r(t) \sin t \Big (r'(t) \cos t - r(t) \sin t \Big) + r(t) \cos t \Big( r'(t) \sin t + r(t) \cos t \Big)} {r^2 (t)} = 1$$

which, integrated from $0$ to $2\pi$, will give $2 \pi$.

Please note that the application of Green's theorem on $\Omega _\varepsilon$ will tell you that the integral on $C$ is equal to the integral on the small circle of radius $\varepsilon$ surrounding the origin, but won't calculate things for you. You'll still have to parametrize the circle and do an explicit integral.