Suppose $T$ is symmetric, and has a unique selfadjoint extension. (I'll assume that $T$ is densely-defined, though you can probably prove that it must be the case.)
Let $\overline{T}$ be the closure of $T$, which exists because $T$ is symmetric. $\overline{T}$ must also be symmetric, and because $\overline{T}^*=T^*$, then
$$
\mathcal{D}(T^*)=\mathcal{D}(\overline{T})\oplus\mathcal{N}(T^*-iI)\oplus\mathcal{N}(T^*+iI),
$$
where the decomposition is orthogonal with respect to the graph inner product
$$
(x,y)_{T^*} = (x,y)_H+(T^*x,T^*y)_{H}.
$$
If $S$ is a selfadjoint extension of $T$, then $S$ is also a selfadjoint extension of $\overline{T}$ because the graph of $T$ is contained in the graph of $S$ and the graph of $S$ is closed. $\overline{T}$ has a selfadjoint extension iff $\mathcal{N}(T^*-iI)$ is unitarily equivalent to $\mathcal{N}(T^*+iI)$, and the possible selfadjoint extensions are in one-to-one correspondence with the unitary maps $U: \mathcal{N}(T^*-iI)\rightarrow\mathcal{N}(T^*+iI)$. However, because $\overline{T}$ has only one selfadjoint extension, then $\mathcal{N}(T^*-iI)=\{0\}=\mathcal{N}(T^*+iI)$ follows. Hence, $\overline{T}=T^*$, which also gives $\overline{T}=\overline{T}^*$. So $\overline{T}$ must be selfadjoint.
The difference is where you are testing your operator.
To see that an operator is symmetric, you proceed as you did and show that given two generic elements of your domain $x,\ y\in\mathcal D\left(D\right)$ you have that $ \langle x, Dy\rangle= \langle Dx,y\rangle $ (with the same $ D $!).
On the other hand, the concept of self-adjoint goes hand in hand with the concept of adjoint; if you have that $ D^* $ is the adjoint of $ D $, then self-adjoint means only that as operators $ D $ and $ D^* $ coincide, i.e., their domains are equal and the action on the domains are equal.
So we are reduced to understand what the adjoint of an operator $ D $ is. To do so, you first assume that your operator $ D $ is densily defined (which is your case) and define the domain
$$
\mathcal D\left(D^*\right):=\left\{x\in\ell^2|\ \exists C\mbox{ s.t. }\left|\langle x,Dy\rangle\right|\le C\left\|y\right\| \forall y\in \mathcal D\left(D\right)\right\}.
$$
Then, for any element in $ \mathcal D\left(D^*\right) $ the mapping $ y\mapsto\langle x,Dy\rangle $ is define on a dense of your Hilbert space $ \ell^2 $ and it is bounded, so it can be extended to an element of the dual, and therefore, via Riesz representation theorem there is a unique element of your space $ z $ such that $ \langle x,Dy\rangle=\langle z,y\rangle $ for any $ y\in\mathcal D\left(D\right) $. By definition, $ D^*x:=z $.
Getting to the exercise. As you noticed, $ \langle e_n, De_m\rangle=n\delta_{nm}=\langle De_n, e_m\rangle $, therefore, given $ x,\ y\in\ell^2 $ we have
\begin{align}
\langle x,D y\rangle
&=\langle\sum_{n\in\mathbb N}x_ne_n,D\sum_{m\in\mathbb N}y_me_m\rangle
=\sum_{n\in\mathbb N}\sum_{m\in\mathbb N}x_n y_m\langle e_n,De_m\rangle
\\
&
=\sum_{n\in\mathbb N}\sum_{m\in\mathbb N}x_n y_mm\delta_{nm}
=\sum_{n\in\mathbb N}\sum_{m\in\mathbb N}x_n y_mn\delta_{nm}
=\langle Dx,y\rangle,
\end{align}
and $ D $ is symmetric.
Now I will show you that $ \mathcal D\left(D^*\right)=\mathcal D\left(D\right) $. The inclusion $ \supseteq $ is trivial, so let's show the other inclusion. Let $ x $ be in $ \mathcal D\left(D^*\right) $. Consider the sequence of elements $ y(k) $ of $ \mathcal D\left(D\right) $ (a sequence of sequences) defined as
$$
y(k)=\sum_{n=0}^k\frac{nx_n}{\left\|\sum_{n=0}^knx_ne_n\right\|}e_n.
$$
With this definition $ \|y(k)\|=1 $ and $ y(k)\in\mathcal D\left(D\right) $ trivially, given that it is a finite sequence. Therefore we get that, given that $ x\in\mathcal D\left(D\right) $, there exists a constant $ C $ such that
\begin{align}
C=C\left\|y(k)\right\|&\ge\left|\langle x,Dy\left(k\right)\rangle\right|
=\left|\sum_{n\in\mathbb N}\overline{x_n}ny(k)_n\right|
=\frac1{\left\|\sum_{n=0}^knx_ne_n\right\|}\left|\sum_{n=0}^kn^2\left|x_n\right|^2\right|
\\
&=\left(\sum_{n=0}^kn^2\left|x_n\right|^2\right)^\frac12.
\end{align}
It follows immediately that we can send $ k $ to infinity and get that the series in the last term is convergent and $ x\in\mathcal D\left(D\right) $ and the two domains coincide.
Finally is an exercise of dialectic to prove that now the actions of $ D $ and $ D^* $ coincide on their (common) domain. Indeed let $ x,y\in\mathcal D\left(D\right)=\mathcal D\left(D^*\right) $, by definition of $ D^* $ and using the symmetry we get
\begin{align}
\langle x,D^* y\rangle
=\langle Dx,y\rangle
=\langle x,D y\rangle.
\end{align}
Given that the domain of $ D $ is dense, this implies $ D^*y=Dy $ for every y in the domain, and therefore $ D=D^* $. Hence, $ D $ is also self-adjoint.
Best Answer
The adjoint $T^*$ is defined as the set of $g\in L^2(\mathbb{R})$ for which there exists a constant $C_{g}$ such that $$ |\langle Tf,g\rangle_{L^2}| \le C_g\|f\|_{L^2},\;\;\; \forall f\in \mathcal{D}(T). $$ This inequality holds iff there is a unique $T^*g\in L^2$ such that $$ \langle Tf,g\rangle = \langle f,T^*g\rangle,\;\;\; \forall f\in\mathcal{D}(T). $$ ($T^*g$ is unique if it exists because $\mathcal{D}(T)$ is dense in $L^2(\mathbb{R})$.) The Fourier transform $\mathcal{F}$ on $L^2$ can be brought to bear on $|\langle Tf,g\rangle| \le C_g\|f\|_{L^2}$: $$ \langle \widehat{Tf},\widehat{g}\rangle=\langle \widehat{f},\widehat{T^*g}\rangle \\ \langle -\xi^2\widehat{f},\widehat{g}\rangle=\langle \widehat{f},\widehat{T^*g}\rangle \\ \langle\widehat{f},-\xi^2\widehat{g}\rangle=\langle \widehat{f},\widehat{T^*g}\rangle \\ \implies \widehat{T^*g}=-\xi^2\widehat{g} \in L^2 \\ T^*g = -\mathcal{F}^{-1}\xi^2\mathcal{F}g $$ So the adjoint $T^*$ is fully characterized in terms of the Fourier transform: it is unitarily equivalent to multiplication by $-\xi^2$ in the Fourier domain. Multiplication operators on $L^2(\mathbb{R})$ are self-adjoint. $$ T^*=-\mathcal{F}^{-1}\xi^2\mathcal{F} \\ \implies T^c = (T^*)^*=-\mathcal{F}^{-1}\xi^2\mathcal{F}, $$ where $T^c$ is the closure of $T$. $T^c$ is self-adjoint because it is unitarily equivalent to a multiplication operator.