Counterexamples abound. Here are a few of them.
Theorem. If $\emptyset\ne X\subseteq\mathbb R$ and $X\subseteq\operatorname{cl}(\operatorname{int}(X))$, then $\mathcal D(X)\cong\mathcal D(\mathbb R)$.
Proof. Construct an infinite sequence of pairwise disjoint open intervals $I_n$ so that $\bigcup_{n=1}^\infty I_n$ is a dense subset of $X$, and $|X\setminus\bigcup_{n=1}^\infty I_n|=2^{\aleph_0}$. Then a set $D\subseteq X$ is dense in $X$ if and only if $D\cap I_n$ is dense in $I_n$ for each $n$.
Thus, if $X'$ is another set satisfying the same hyotheses, with an analogous sequence of intervals $I'_n$, then a bijection $f:X\to X'$ which maps each $I_n$ homeomorphically onto the corresponding $I'_n$ is an isomorphism from $\mathcal D(X)$ to $\mathcal D(X')$.
P.S. With a somewhat more complicated argument one can prove:
Theorem. If $X$ and $Y$ are nonempty Polish spaces with no isolated points, then $\mathcal D(X)\cong\mathcal D(Y)$.
The idea is to construct a family $(U_i:i\in I)$ of open subsets of $X$ and a family $(V_i:i\in I)$ of open subsets of $Y$ so that:
(1) a set $D\subseteq X$ is dense in $X$ if and only if $D\cap U_i\ne\emptyset$ for each $i\in I$;
(2) a set $D\subseteq Y$ is dense in $Y$ if and only if $D\cap V_i\ne\emptyset$ for each $i\in I$;
(3) there is a bijection $f:X\to Y$ such that $f[U_i]=V_i$ for each $i\in I$.