The Poincaré inequality need not hold in this case. The region where the function is near zero might be too small to force the integral of the gradient to be large enough to control the integral of the function.

For an explicit counterexample, let $$\Omega = \{(x,y) \in \mathbb{R}^2 : 0 < x < 1, 0 < y < x^2\}$$ be the region under the graph of a parabola, and take $p=2$. For $\epsilon > 0$, let $$u_\epsilon(x,y) = \begin{cases} x/\epsilon, & x < \epsilon \\\ 1, & x \ge \epsilon \end{cases}.$$ $0$ is in the essential range of each $u_\epsilon$, but one can easily verify $||\nabla u_\epsilon||_2^2 = \epsilon/3 \to 0$ as $\epsilon \to 0$, whereas $||u_\epsilon||_2^2 \to |\Omega| = 1/3$.

**Edit:** To address Piero's comment, the irregularity of the domain is not the main issue here. For another counterexample, let $\Omega$ be the unit ball in $\mathbb{R}^d$, $d \ge 3$, and take
$$u_\epsilon(x) = \begin{cases} \frac{|x|^2}{\epsilon^2}, & |x| < \epsilon \\\ 1, & |x| \ge \epsilon \end{cases}.$$ Using polar coordinates, one easily computes $||\nabla u_\epsilon||_2^2 \sim \epsilon^{d-2}$ while again $||u_\epsilon||_2^2 \to |\Omega|$.

What's really the problem is that the set where $u$ vanishes has zero capacity. If you can control from below the capacity of the set where $u$ vanishes, then you can get a Poincaré inequality.

Indeed, the following theorem may be what the asker wants: for "reasonable" $\Omega$ (Lipschitz suffices), if $\mathrm{Cap}(\{u = 0\}) \ge \delta$, then $||u||_2^2 \le \frac{C}{\delta} ||\nabla u||_2^2$. The precise statement (for any value of $p$) can be found in section 4.5 of William P. Ziemer's *Weakly Differentiable Functions* (along with the definition of capacity).

Note in particular that $\mathrm{Cap}(E) \ge m(E)$, so it's enough to control the Lebesgue measure. This is your bullet point 3.