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.
A necessary and sufficient condition for $\ A^*\ $ to be positive is that $\ A^*=A^{**}\ $—that is, the adjoint of $\ A\ $ is self-adjoint. In fact, $\ A\ $ must be essentially self-adjoint, and $\ A^*\ $ must be the unique self-adjoint extension of $\ A\ $.
If $\ A^*\ $ is self-adjoint, then its spectrum is a subset of the real line, and the operators $\ A^*+iI\ $ and $\ A^*-iI\ $ therefore have bounded inverses.
It follows that $\ \ker\big(A^*+iI\big)= \ker\big(A^*-iI\big)=\{0\}\ $, and hence that
\begin{align}
\mathscr{R}(A+iI)^\perp&=\ker(A^*-iI)=\{0\}\\
\mathscr{R}(A-iI)^\perp&=\ker(A^*+iI)=\{0\}\ .
\end{align}
Thus, $\ A\ $ is essentially self-adjoint because both its deficiency indices are zero. Since $\ A\ $ therefore has a unique self-adjoint extension, its adjoint, its Friedrichs extension and its Krein-von Neumann extension must all coincide. The adjoint must therefore be positive.
On the other hand, if $\ A^*\ $ is positive, then it is symmetric, and so $\ A^{**}\supseteq A^*\ $. Also, if $\ x\in\mathscr{R}(A+iI)^\perp=$$\,\ker(A^*-iI)\ $, then
$$
\langle A^*x,x\rangle=i\|x\|^2\ ,
$$
which implies that $\ x=0\ $ because $\ A^*\ $ is positive. Therefore,
$$
\mathscr{R}(A+iI)^\perp=\{0\}\ ,
$$
and the deficiency indices of $\ A\ $ (which must be equal because $\ A\ $ is positive symmetric) are zero. So again $\ A\ $ is essentially self-adjoint. Let $\ B\ $ be the unique self-adjoint extension of $\ A\ $. Since $\ B\supseteq A\ $, then $\ B=B^*\subseteq A^*\ $, and therefore, since $\ B\ $ is symmetric, $\ B^*\supseteq A^{**}\ $. We now have the chain of extensions
$$
\ A^{**}\supseteq A^*\supseteq B\supseteq A^{**}\ ,
$$
which imply that
$$
A^{**}=A^*=B
$$
and hence that $\ A^*\ $ is self-adjoint.
That essential self-adjointness isn't sufficient by itself is shown by the following example of an essentially self-adjoint positive operator, defined on a dense subspace of $\ \mathscr{L}^2[0,1]\ $, whose adjoint is not positive.
Define
$$
(Af)(x)=-f''(x)
$$
for all twice-differentiable functions $\ f:[0,1]\rightarrow\mathbb{C}\ $ satisfying $\ f'(0)=f'(1)=f(0)=f(1)=0\ $ and $\ \int_0^1|f''(x)|^2\,dx<$$\,\infty\ $. This space of functions is dense (with respect to the $\ \mathscr{L}^2\ $ norm) in $\ \mathscr{L}^2[0,1]\ $. Integration by parts gives
\begin{align}
\langle Af,f\rangle&=-\int_0^1f''(x)\overline{f(x)}\,dx\\
&=\int_0^1|f'(x)|^2\,dx\\
&\ge0
\end{align}
for all $\ f\in\mathscr{D}(A)\ $, so $\ A\ $ is positive. Now if $\ g:[0,1]\rightarrow\mathbb{C}\ $ is defined to be
$$
g(x)=x^2\ ,
$$
then $\ g\in\mathscr{L}^2[0,1]\ $ and
\begin{align}
\langle Af,g\rangle&=-\int_0^1f''(x)x^2\,dx\\
&=2\int_0^1f'(x)x\,dx\\
&=-2\int_0^1f(x)\,dx\\
&=\langle f,h\rangle
\end{align}
for all $\ f\in\mathscr{D}(A)\ $, where $\ h(x)=-2\ $ for $\ x\in[0,1]\ $. It follows that $\ g\in\mathscr{D}\big(A^*\big)\ $ and $\ A^*g=h\ $. Now
\begin{align}
\langle A^*g,g\rangle&=\int_0^1-2x^2\,dx\\
&=-\frac{2}{3}\ ,
\end{align}
and therefore $\ A^*\ $ is not positive.
The essential self-adjointness of $\ A\ $ follows from the following properties:
\begin{align}
\mathscr{R}(A+iI)^\perp&=\{0\}\\
\mathscr{R}(A-iI)^\perp&=\{0\}\ ,
\end{align}
which are not difficult to prove, and imply that both deficiency indices of $\ A\ $ are $\ 0\ $.
Best Answer
Suppose that $A^*$ and $B^*$ are both adjoints for the operator $A$. Then, $A^* \psi = \eta$ and $B^* \psi = \tilde \eta$. We then have, from the definition of the adjoint and sesquilinearity of the inner product, $$\langle \tilde \eta, \alpha\rangle = \langle \eta, \alpha\rangle\implies \langle\tilde \eta- \eta, \alpha\rangle = 0, \forall\alpha\in \mathcal D_A. $$ Now, a set $D$ is dense in the Hilbert space $\mathcal H$ if $\bar D = \mathcal H.$ There are several equivalent definitions of the topological closure, but the most useful one for our purposes is the following: $\bar D$ is the set consisting of $D$ and all of its limit points. If a point $x \in \mathcal H$ is a limit point for $D$, then there is a sequence $\{x_n\}_{n\in\mathbb N}$ in $D-\{x\}$ converging to $x$ (this is because every metric space, and thus every Hilbert space, is a Fréchet–Urysohn space). Now, since $\mathcal {\bar D}_A = \mathcal H,$ every point in $\mathcal H$ is the limit of some sequence in $\mathcal D_A$. Let $\{x_n\}_{n\in \mathbb N}$ be such a sequence converging to $\tilde \eta - \eta\in \mathcal H.$ By the above arguments, $$\langle \tilde \eta - \eta, x_n\rangle = 0. $$ It can be shown that the inner product is sequentially continuous, that is if $\lim_{n\to \infty} \phi_n = \phi$, then $\lim_{n\to \infty} \langle\psi, \phi_n\rangle = \langle \psi , \phi\rangle$ for all $\psi \in \mathcal H$. This implies that $$\langle \tilde \eta - \eta, \tilde \eta - \eta\rangle = 0 \implies \tilde \eta = \eta$$ by positive definiteness. Thus, the adjoint is unique.