If $a(x) > 0$ instead of $a(x) \ge 0$, case 2 can be handled like this:
Since $x_0$ is a local maximum of $w$, we have
$$\left.\frac{dw}{d\nu}\right|_{x=x_0} \ge 0 \implies
-a(x_0)w(x_0) \ge 0 \implies w(x_0) \le 0$$
From this, it follows $w(x) \le 0$ over $\Omega$. The rest follows essentially same argument in your analysis of case 1.
I have no idea what happens if $a(x_0)$ is allowed to vanish. However, there is an alternate way to prove the uniqueness. In fact, we can generalize it a little bit.
Let $F : \Omega \times \mathbb{R} \to \mathbb{R}$ be any regular enough function such that for any fixed $x$, $F(x,y)$ is strictly increasing in $y$. For the problem at hand, take $F(x,y)$ to be $y^3$.
Let $u, v$ be two solutions of the boundary value problem:
$$\begin{array}{ll}
-\Delta \phi(x) + F(x,\phi(x)) = 0,& x \in \Omega\\
\hat{n}(x)\cdot \nabla\phi(x) + a(x) \phi(x) = h(x),& x \in \partial\Omega
\end{array}
$$
where $\hat{n}(x)$ is the unit outward normal of $\partial\Omega$ at $x$.
Let $w = u-v$ be their difference. Since $F(x,y)$ is increasing in $y$, for any $x \in \Omega$, we have
$$w(x)(F(x,u(x))-F(x,v(x)) \ge 0$$
Furthermore, equality is achieved at and only at those $x$ where $w(x) = 0$.
From this, we find
$$\mathcal{I} \stackrel{def}{=} \int_\Omega (|\nabla w(x)|^2 + w(x)(F(x,u(x))-F(x,v(x))) dx \ge 0$$
Since $|\nabla w|^2 = \nabla\cdot( w \nabla w) - w \Delta w$, the integrand above can be simplified to $\nabla\cdot(w \nabla w)$.
Apply divergence theorem, we obtain
$$\mathcal{I} = \int_\Omega \nabla\cdot (w \nabla w) dx =
\int_{\partial\Omega} \hat{n}\cdot (w \nabla w) dS_x
= -\int_{\partial\Omega} aw^2 dS_x \le 0$$
This forces $\mathcal{I} = 0$ and hence for all $x \in \Omega$:
$$\begin{align}
& |\nabla w(x)|^2 + w(x)(F(x,u(x))-F(x,v(x))) = 0 \\
\implies & w(x)(F(x,u(x))-F(x,v(x))) = 0 \\
\implies & w(x) = 0
\end{align}$$
The proof I am aware of (Ian's comment) of Strong Maximum Principle does not use boundedness.
But anyway, suppose the Strong Maximum Principle only for bounded, open, connected sets. Let $\Omega$ be unbounded, open, and connected, with $u(x_0)=\max_{\overline \Omega}u$ for some $x_0\in\Omega$. Open connected in $\mathbb R^n$ is path connected. Let $\Omega_n$ be the open, connected set of all points in $\Omega$ which can be joined to $x_0$ by a path of length $< n$ . Now we can apply the "SMP for bounded sets" on each $\Omega_n$ to deduce $u|_{\Omega_n} \equiv u(x_0)$. As $\bigcup_{n=0}^\infty \Omega_n = \Omega$, this proves SMP for unbounded sets.
For your reference I copy the standard proof of SMP from my notes:
Theorem (Strong maximum principle). Let $\Omega$ be open and connected (possibly unbounded), and let $u$ be subharmonic in
$\Omega$. If u attains a global maximum value in $\Omega$, then $u$ is
constant in $\Omega .$
Proof. Define $M:=\max _{\Omega} u$, and $A:=u^{-1}\{M\}=\{x \in
\Omega: u(x)=M\}$. By the assumption of the theorem, $A$ is non-empty.
Since $u$ is continuous, it follows that $A$ is relatively closed in
$\Omega$ (i.e., there exists a closed $F \subset \mathbb{R}^{n}$ such
that $\left.A=F \cap \Omega\right)$. We now show that $A$ is also
open. Indeed, let $x \in \Omega$ and let $r$ be sufficiently small
that $\overline{B_{r}(x)} \subset \Omega$. By the mean value property
for subharmonic functions, we have $$ 0=u(x)-M \leq
\frac{1}{\left|B_{r}(x)\right|} \int_{B_{r}(x)}(u(y)-M) d y $$ Since
$M=\max _{\Omega} u$, we have $u(y)-M \leq 0$ for all $y \in
B_{r}(x)$. This implies that $$ 0=u(x)-M \leq
\frac{1}{\left|B_{r}(x)\right|} \int_{B_{r}(x)}(u(y)-M) d y \leq 0 $$
Since $u$ is continuous, this can only happen if $u=M$ in $B_{r}(x)$.
So that $B_{r}(x) \subset A$, which implies that $A$ is open. Since
$\Omega$ is connected and $A \neq \emptyset$ is both open and closed,
it follows that $A=\Omega$, and hence $u \equiv M$ in $\Omega$. This
finishes the proof.
Best Answer
We calculate \begin{align*} \frac{\partial u}{\partial x_i} &= \frac{\partial }{\partial x_i} \sum_{k=1}^2 \bigg ( \frac{\partial v}{\partial x_k} \bigg )^2 \\ &= 2 \sum_{k=1}^2 \frac{\partial v}{\partial x_k} \cdot \frac{\partial^2 v}{\partial x_k\partial x_i} \end{align*} and \begin{align*} \frac{\partial^2 u}{\partial x_i^2} &= 2 \sum_{k=1}^2 \bigg [ \bigg (\frac{\partial^2 v}{\partial x_k\partial x_i} \bigg )^2 +\frac{\partial v}{\partial x_k} \cdot \frac{\partial^3 v}{\partial x_k\partial x_i^2}\bigg ]\\ &\geqslant 2 \sum_{k=1}^2 \frac{\partial v}{\partial x_k} \cdot \frac{\partial^3 v}{\partial x_k\partial x_i^2} \end{align*} It follows that $$ \Delta u \geqslant 2 \sum_{k=1}^2 \frac{\partial v}{\partial x_k} \cdot \frac{\partial }{\partial x_k} \Delta v =0 $$ since $\Delta v$ is constant. Hence, $u$ is subharmonic. Then the maximum principle implies $u$ attains its maximum on the boundary.
*Note that this argument requires $v \in C^3(\Omega)$. This is not a big deal since standard regularity theory implies that $v$ is in fact smooth in $\Omega$.