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The answer is yes: take any smooth function $g_0$ on $\partial D_1$ and extend it to a smooth function $g$ on $\mathbb{R}^n$. Multiply by a smooth cutoff function $\eta$ which is $1$ near $\partial D_1$ and $0$ near $\partial D_2$. Then $f = \Delta (\eta g)$ is smooth, so in particular in $L^2$. This shows that $g_0$ is in the image of your operator $S$, and smooth functions are dense in $H^s$.