First we look for a distributional solution. Remember that, as an distribution, $\Delta u$ is defined by $$\langle\Delta u,v \rangle=-\langle\nabla u,\nabla v\rangle,\ \forall\ v\in C_0^\infty(\Omega). \tag{1}$$
From $(1)$, we can say that a solution in the distributional sense, is a function $u\in H^1(\Omega)$ with $Tu=g$ satisfying $$\int_\Omega \nabla u\nabla v=0,\ \forall\ v\in C_0^\infty(\Omega). \tag{2}$$
By density we may conclude from $(2)$ that $$\int_\Omega \nabla u\nabla v=0,\ \forall v\in H_0^1(\Omega). \tag{3}$$
This is not the only weak formulation for this problem, however, it is the one which comes from a variational problem, to wit, let $F:\{u\in H^1(\Omega):\ Tu=g \}\to \mathbb{R}$ be defined by $$Fu=\frac{1}{2}\int_\Omega |\nabla u|^2.$$
Note that $(3)$ can be rewritten as $$\langle F'(u),v\rangle=0,\ \forall\ v\in H_0^1(\Omega). \tag{4}$$
Also note that, once $\{u\in H^1(\Omega):\ Tu=g \}$ is a closed convex set of $H^1(\Omega)$ and $F$ is a coercive, weakly lower semi continuous function, we have that $F$ has a unique global minimum which satisfies $(4)$.
By not considering any kind of derivative of $u$, you can also use another weak formulation: let $C_0^{1,\Delta}(\overline{\Omega})=\{u\in C_0^1(\overline{\Omega}):\ \Delta u \in L^\infty(\Omega)\}$. A "very" weak solution, is a function $u\in L^1(\Omega)$ satisfying $$\int_\Omega u\Delta v=-\int_{\partial\Omega}g\frac{\partial v}{\partial \nu },\ \forall\ v\in C_0^{1,\Delta}(\overline{\Omega}).$$
In your setting, I mean, when $g\in H^{1/2}(\Omega)$, it can be proved that both definitions are equivalent. For references, take a look in the paper Elliptic Equations Involving Measures from Veron. It has a PDF version here. Take a look in page 8.
To conclude, I would like to adress @JLA, which gave a comment in OP; in the end, what we really want is a $H^2$ function (or more regular), because we are working with the Laplacean and it is natural to have two derivatives.
It can be proved, by using regularity theory, that $u$ is in fact in $H^2$, however, there is a huge difference between proving that $u$ is in $H^2$ after finding it in $H^1$ by the above methods and finding directly $u\in H^2(\Omega)$ by another method. Note, for example, that none of the methods above, does apply if we change $H_0^1(\Omega)$ by $H^2(\Omega)$.
Best Answer
First, a proof of the Poincaré inequality for domains of bounded width is given here: poincaré inequality direct proof
You are absolutely right that this implies that the Poisson problem has a unique solution for all $f\in L^2(\Omega)$ if the domain $\Omega$ has bounded width. The usual proof using the Lax-Milgram lemma goes through without change.
Here is a different way to think of it using spectral theory: One can easily verify that the $L^2$ Poincaré inequality is equivalent to a spectral gap of the Dirichlet Laplacian, i.e., the existence of $\lambda_0>0$ such that $\sigma(\Delta^{(D)})\subset [\lambda_0,\infty)$. Of course this means that $\Delta^{(D)}$ is invertible, i.e., the Poisson problem has a unique solution.
This spectral gap is not directly related to the discreteness of the spectrum, it may be continuous spectrum that just starts at $\lambda_0$. However, if you already know that $\Delta$ has purely discrete spectrum, then it's easy to check whether it has a spectral gap - you just have to check if $0$ is an eigenvalue, and since an eigenfunction for $0$ satisfies $\int |\nabla f|^2=0$, this boils down to checking whether the constant functions are in the domain of your particular realization of the Laplacian.