dual of $H^1_0$: $H^{-1}$ or $H_0^1$?

Let me restate your question as follows. Let $\Omega$ be a nonempty open subset of $\mathbb{R}^d$. On the one hand, we have the sandwich: $$ H_0^1(\Omega)\subset L^2(\Omega)\subsetneq H^{-1} (\Omega),\tag{*} $$ where $H^{-1}(\Omega)$ is defined as the dual of $H_0^1(\Omega)$. On the other hand, by Riesz representation theorem, $H^{-1}(\Omega)$ can be identified as $H_0^1(\Omega)$. This suggests that we may write $$ H^{-1}(\Omega)=H_0^1(\Omega)\tag{**} $$ which contradicts $(*)$ since $L^2(\Omega)$ is a proper subspace of $H^{-1}(\Omega)$.

The problem is that $(**)$ is not true if one considers "$=$" in $(**)$ as equality of sets rather than an isomorphism of Hilbert space.

As Terry Tao said:

"It is better to think of isomorphic pairs as being equivalent or identifiable rather than identical, as the latter can lead to some confusion if one treats too many of the equivalences as equalities. For instance, $\ell^2(\{0,1,2,\dots\})$ and $\ell^2(\{1,2,\dots\})$ are equivalent (one can simply shift the standard orthonormal basis for the former by one unit to obtain the latter), but one can also identify the latter space as a subspace of the former. It is fairly harmless to treat one of these equivalences as an equality, but of course one cannot do so for both equivalences at the same time."

The dual space $(H_0^1)^*$ is per definition the space of all linear and continuous functionals $(H_0^1)^*\to \mathbb R$.

There are different ways in generating this dual space:

(1) Riesz representation lets you represent each functional $f\in (H_0^1)^*$ by scalar product with an element $u_f\in (H_0^1)$ $$ f(v) = \langle u_f,v\rangle \quad\forall v\in H_0^1. $$

(2) Every function $u\in L^2$ also generates a function in $(H_0^1)^*$ via $$ f_u(v)=\int_\Omega uv\ dx \quad\forall v\in H_0^1. $$ You can even take $u\in H_0^1$.

If you look at the different functionals, then you see that the functionals in (1) are defined by the $H^1$-scalar product, the functionals in (2) are defined by the $L^2$-scalar product.

Riesz representation theorem is a : "representation" theorem, and not an identification theorem. In the example 3 above, $f = - \Delta u + u$ can be represented by $u$, but cannot be identified by $u$, since $f$ is in $H^{-1}$ whereas $u$ is in $H^1_0$.