$\require{extpfeil}\Newextarrow{\xRightarrow}{5,5}{0x21D2}$
Proof of $(\Rightarrow)$:
Proving that $A^*$ is monotone.
Assume that $A$ is maximal monotone. Then, for all $f \in H$ there exists $u \in D\left(A\right)$, such that $u + Au = f$. Let $f = A^*u$, which yields:
$$u+ Au = A^* u \Leftrightarrow Au = A^* u - u$$
Obviously, $A$ is also monotone, thus:
$$\left\langle Au, u\right\rangle \geq 0 \Rightarrow \left\langle A^* u-u, u\right\rangle \geq 0 \Leftrightarrow \left\langle A^*u,u \right\rangle - \left\|u\right\|^2 \geq 0 \Leftrightarrow \left\langle Au, u\right\rangle \geq \left\|u\right\|^2.$$
Thus, it is $\left\langle A^*u,u\right\rangle \geq 0$ for all $u \in D\left(A\right) \cap D\left(A^*\right)$. Now, assume that $v \in D\left(A^*\right)$ with $v \not\in D\left(A\right)$ and $f \in H^*=H$. Then, for $u \in D\left(A\right)$, it is:
$$\left\langle Au - f, u-v\right\rangle < 0.$$
Let now $f = -A^*v$ and yield:
\begin{align*}
\left\langle Au + A^*v, u-v\right\rangle <0 &\Leftrightarrow \left\langle Au, u\right\rangle - \left\langle A^*v,v\right\rangle <0 \\&\Leftrightarrow \left\langle A^*v,v\right\rangle > \left\langle Au,u\right\rangle \geq 0.
\end{align*}
So, we showed that $\left\langle A^*v,v\right\rangle \geq 0$ for all $v \in D\left(A^*\right)$, which means that $A^*$ is monotone.
Proving that $D\left(A\right)$ is dense in $H$.
Let $h \in H$ such that $\langle h,u\rangle = 0$ for all $u \in D(A)$. To show that $D(A)$ is dense in $H$, it suffices to show that $h=0$. Since $A$ is maximal monotone, there exists $u_0 \in D(A)$ such that $u_0 + Au_0 = h$. Then, by our initial hypothesis, it will be $(h,u_0) = 0$. But:
$$\langle h,u_0\rangle = \|u_0\|^2 + \langle Au_0,u_0\rangle \geq \|u_0\|^2 \xRightarrow{\langle h,u_0\rangle = 0} u_0 = 0$$
Thus, it will be $h = 0$.
Proving that $A$ is closed.
Initially, we will show that for all $h \in H$ there exists a unique $u \in D(A)$, such that $u + Au = h$. Indeed, let $\bar{u}$ be another solution. Then, we have that:
$$(u-\bar{u}) + A(u-\bar{u}) = 0 \Rightarrow \|u-\bar{u}\|^2 + \langle A\left(u-\bar{u}\right),u-\bar{u}\rangle = 0.$$
But, obviously, $A$ is also monotone, thus: $$\langle A\left(u-\bar{u}\right),u-\bar{u}\rangle \geq 0.$$
That means that: $$\|u-\bar{u}\|^2 = 0 \Leftrightarrow u - \bar{u} = 0.$$
Now, it is:
$$\|u\|^2 + \langle Au,u \rangle = \langle h,u\rangle \Rightarrow \|u\|^2 \leq \langle h,u \rangle \implies \|u\| \leq \|h\|.$$
Therefore, the operator $h \mapsto u$ under the notation $(I + A)^{-1}$, is a bounded linear operator from $H$ to $H$, with $\|(I+A)^{-1}\|_{\mathcal{L}(H)} \leq 1$.
In order now to show that $A$ is closed, let $\{u_n\}_{n=1}^\infty \subset D(A)$ with $u_n \to u$ and $Au_n \to h$. It is true that $u_n + Au_n \to u+ h$, thus it is:
$$u_n = (I+A)^{-1}(u_n + Au_n) \to (I+A)^{-1}(u+h).$$
But $u_n \to u$ and consequently the two limits must coincide, which means that $u = (I+A)^{-1}(u+h)$. We thus conclude that $u \in D(A)$ and $u + Au = u + h$, meaning that $A$ is closed.
Proof of $(\Leftarrow)$:
We now assume that the adjoint operator $A^*$ is monotone and since $A$ is closed, then $A = \left(A^*\right)^*$. As such, we can use the following theorem, stated by Brezis and Browder in the book "Nonlinear Analysis: A collection of papers in honor of Erich H. Rothe.
\textbf{Theorem (Brezis-Browder):} Let $X$ be a reflexive Banach space and $L_0, L_1$ two linear monotone mappings from $X$ to $2^{X^*}$, such that $L_0 \subseteq L_1^*$. Then there exists a monotone linear mapping $L$, such that:
$$L_0 \subseteq L \subseteq L_1^*.$$
Now, from the theorem stated, we can prove that there exists a linear monotone operator $A'$, such that:
$$A \subseteq A' \subseteq \left(A^*\right)^* = A.$$
This means that $A=A'$ and thus $A$ will be a maximal monotone operator.
