I agree with motivations A and B you gave in the text. Even more disappointingly, I would say that one simply chooses a Hilbert space instead of another because it works. So one chooses the least possible regularity and the simplest way to encode boundary conditions.
In the case of the Dirichlet problem, for example:
$$\tag{(D)} \begin{cases} -\Delta u = f & \Omega \\ u= 0 & \partial \Omega\end{cases},$$
a solution $u$, whatever it is, must be something that realizes
$$b(u, v)=0,\quad \forall v \in \text{some test function space}, $$
where
$$b(u, v)=\int\left(-\Delta u - f\right)v\, dx, $$
whenever this makes sense. Turns out that, if we require $u, v\in H^1_0(\Omega)$, then $b$ takes on a super-nice form, the Lax-Milgram's theorem kicks in, everything goes on smoothly and our lives are beautiful. What if we had taken $H^2_0$ instead? Well, in this case we would have had trouble because, even in the simplest case $f=0$, the quadratic form
$$b(u,u)=\int \lvert \nabla u \rvert^2\, dx $$
is not coercive, because it cannot control second derivatives. So the Lax-Milgram's theorem doesn't apply and our lives are miserable.
(A last remark which may possibly contradict everything above. As far as I know, there is an abstract theory of linear operators and quadratic forms on Hilbert spaces which, among other things, proclaims that $H^1_0$ is the "right" domain for the quadratic form $b(u,u)$ when $L$ is the Laplacian. If you really are interested in this you could look for the keywords "form domain of self-adjoint operators" or "Friedrichs extension". I am sure that those things are treated in Reed & Simon's Methods of Modern Mathematical Physics and in Zeidler's Applied Functional Analysis.)
EDIT: This answer is related to the last remark. The book by Davies explains this abstract theory IMHO very clearly.
P.S.: The dichotomy "our lives are beautiful / our lives are miserable" is a citation of J.L. Vázquez.
Let $H=H_0^1\cap H^2$. A good norm to work with is $$\|u\|_H=\|\Delta u\|_2$$
As you can see here, this norm is equivalently to the usual one. Also, the inequality you are looking for is true, in fact, you have that $$\int |\Delta u|^2\geq\lambda_1^2\int|u|^2,\ \forall\ u\in H$$
where $\lambda_1$ is the firt eigenvalue associated with $(-\Delta, H_0^1)$. To prove this inequality, note that
\begin{eqnarray}
\int |\nabla u|^2 &=& -\int u\Delta u \nonumber \\
&\leq& \|u\|_2\|\Delta u\|_2 \nonumber
\end{eqnarray}
From the last inequality, you can conclude by using Poincare inequality. Now you can easily apply Lax-Milgram.
Best Answer
The mistake is in the line "But it must be the same solution since $u$ is unique". This holds in your chosen subspace, but does not hold between subspaces. The unique solution in $H_0^1(\Omega)$ is not in general the same solution as in $H$. So there is a unique solution $u_{0}\in H_0^1(\Omega)$ for the problem formulated in $H_0^1(\Omega)$, and a unique solution $u\in H$ for the problem formulated in $H$.