I would like to understand whether the spaces $H^1(\mathbb{R}^3)$ and $H^1_0(\mathbb{R}^3\!\setminus\!\{0\})$ are the same or not.
The first space is the standard Sobolev space, for which one also has $H^1(\mathbb{R}^3)=\overline{C^\infty_0(\mathbb{R}^3)}^{\|\,\|_{H^1}}$, that is, the closure in the $H^1$-norm of the smooth functions compactly supported away from the origin.
The second space is, by definition, $H^1_0(\mathbb{R}^3\!\setminus\!\{0\})=\overline{C^\infty_0(\mathbb{R}^3\!\setminus\!\{0\})}^{\|\,\|_{H^1}}$.
Clearly, $H^1_0(\mathbb{R}^3\!\setminus\!\{0\})$ is a closed subspace of $H^1(\mathbb{R}^3)$.
Thus, an equivalent version of the question is: is the space $C^\infty_0(\mathbb{R}^3\!\setminus\!\{0\})$ dense in $H^1(\mathbb{R}^3)$ ?
Yes, it suffices to show that for every $C^\infty_0(\mathbb{R}^3)$ function $f$ there is a $C^\infty_0(\mathbb{R}^3\setminus \{0\})$ function $g$ such that
$$ \|f - g\|_{H^1} < \delta $$
Take $\chi$ a smooth bump function equaling $1$ on $B_1(0)$ and $0$ outside $B_2(0)$, bounded between $0$ and $1$ everywhere, and has bound $|\partial\chi| < 2$ pointwise everywhere. Let $\chi_\epsilon(x) = \chi(x / \epsilon)$. So that $|\partial\chi_\epsilon| < 2 / \epsilon$.
Let $g = (1-\chi_\epsilon) f$. Then
$$ \|g - f\|_{H^1} = \|\chi_\epsilon f\|_{H^1} \leq \|\chi_\epsilon f\|_{L^2} + \|(\partial\chi_\epsilon) f\|_{L^2} + \|\chi_\epsilon (\partial f)\|_{L^2} $$
As $\epsilon \to 0$ the first and third terms are bounded by $$ C \epsilon^{3/2} \|f\|_{H^1} $$ for some universal constant $C$. The middle term is bounded by $$ C' \epsilon^{1/2} \|f\|_{L^2} $$ using that $$ \|\partial \chi_{\epsilon} f\|_{L^2} \leq \|\frac{2}{\epsilon} f\|_{L^2(B_{2\epsilon}(0))}. $$
So taking $\epsilon$ sufficiently small we can guarantee $\|f - g\|_{H^1} < \delta$.
The argument fails to work in $H^2$: by Sobolev embedding we have $$ \|f - g\|_{L^{\infty}} \leq \|f - g\|_{H^2} $$ so if $f \neq 0$ at the origin, it cannot be approximated by any $C^\infty_0(\mathbb{R}^3\setminus \{0\})$ function.