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.
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 $H$.
Best Answer
The argument is basically the same as a proof of the Poincare inequality in $H_0^1$: Assume that $a$ (as in Davide's comment) is not coercive, then we can take $u_n\in H^1$ with $\| u_n\|_{H^1}=1$ such that $$\| \nabla u_n\|_{L^2(\Omega)}^2 + \| u_n \|_{L^2(\partial\Omega)}^2 \leq n^{-1} \| u_n\|_{H^1}^2=n^{-1}$$
First notice that the inequality gives that $\nabla u_n, u_n|_{\partial\Omega} \to 0$ strongly in the corresponding $L^2$ spaces. By basic results on Sobolev spaces, if we can prove that there is $u\in L^2(\Omega)$ with $u_n \to u$ strongly in this last space, we could conclude two things: $u\in H^1$ and $u$ is constant on each connected component of $\Omega$. So $u$ is locally constant, by smoothness of the domain each connected component must have a nontrivial piece of the boundary, so if $u\neq 0$ in a component we would have a non-zero trace, which is absurd since the trace is a bounded linear operator and $u_n|_{\partial\Omega} \to 0$. We conclude that $u=0$ on $\Omega$. This is of course a contradiction since $u_n\to u$ in $H^1$ and $\|u_n\|_{H^1}=1$.
Now all we have to do is show such an $u$. By reflexivity there is $u\in H^1$ with (really a subsequence) $u_n\rightharpoonup u$ in $H^1$, and by Rellich's theorem $u_n \to u$ in $L^2$.