Let $D_1$ be a domain with smooth boundary and assume that $D_1$ is a proper subset of $D_2$ which is itself a bounded domain in $\mathbb R^n$ with a smooth boundary. Assume also that $D_2\setminus D_1$ is connected. We write $L^2(D_2\setminus D_1)$ for the set of functions in the space $$\{f \in L^2(D_2)\,:\,\textrm{supp}(f)\subset D_2\setminus \overline{D_1}\}$$ let us define the mapping $$S: L^2(D_2\setminus \overline{D_1})\mapsto H^{\frac{3}{2}}(\partial D_1),$$ through $$ Sf:= u_{\partial D_1},$$ where $u \in H^2(D_2)$ is the unique solution to the equation $$ \Delta u =f \quad \text{on $D_2$},$$ subject to $u_{\partial D_2}=0$. Is it true that the image of $S$ is dense in $H^{\frac{1}{2}}(\partial D_1)$?
The answer is yes: take any smooth function $g_0$ on $\partial D_1$ and solve the Dirichlet problem $$ \begin{cases} \Delta g = 0 & \text{ on } D_1\\ g = g_0 & \text{ on } \partial D_1. \end{cases} $$ Now extend $g$ to a smooth function on $\mathbb{R}^n$. Multiply by a smooth cutoff function $\eta$ which is $1$ on $D_1$ and compactly supported on $D_2$.
Then $f = \Delta (\eta g)$ is smooth and supported on $D_2 \setminus \bar{D}_1$, so in particular lies in the given $L^2$ space. This shows that $g_0$ is in the image of your operator $S$, and smooth functions are dense in $H^s$.

$\begingroup$ But why is ${\rm supp}(f)\subset D_2\backslash\overline{D_1}$? $\endgroup$ Oct 5 '21 at 2:02

$\begingroup$ You are correct and I misread part of the question. However, I believe essentially the same argument still applies (solve for the harmonic function first, then extend), and I have modified the answer. $\endgroup$ Oct 5 '21 at 3:03
The answer is yes. Suppose $g$ is orthogonal to the image of $S$, and let $v$ be the solution of the Dirichlet problem $\Delta v=g\delta(\partial D_1)$ on $D_2$, where $\delta(\partial D_1)$ is a delta function localized on $\partial D_1$. We find $$\int_{\partial D_1} gu\,dS=\int_{D_2}v\Delta u\,dx.$$ Now suppose this is true for every $u$ for which $\Delta u$ has compact support in any subregion of $D_2\backslash \overline{D_1}$. Then $v$ must vanish on that subregion. Since $\Delta v=0$ on $D_2\backslash \overline{D_1}$, it follows that $v$ also vanishes there. But $v$ is continuous across $\partial D_1$ (only the normal derivative has a jump), and $\Delta v=0$ in $D_1$, so $v$ must be zero everywhere by uniqueness of the Dirichlet problem for $D_1$. Hence $g=0$.