If $A,B$ are positive, commuting operators, then $AB$ is positive. This is because the unique positive $\sqrt{A}$ must also commute with $B$ and, hence,
$$
\langle ABx,x\rangle = \langle \sqrt{A}Bx,\sqrt{A}x\rangle =
\langle B\sqrt{A}x,\sqrt{A}x\rangle \ge 0.
$$
This result is useful in what follows.
Suppose $A$ is selfadjoint. Let $P=\frac{1}{2}(|A|+A)$ and $N=\frac{1}{2}(|A|-A)$, where $|A|$ is the unique positive square root of $A^2$. Then $PN=NP=0$ and $A=P-N$. This is the desired decomposition of $A$, and the trick is to show that $P,N$ are positive operators.
Let $E$ be the orthogonal projection onto $\mathcal{N}(|A|+A)$. Then $(|A|+A)E=0$ gives $E(|A|+A)=0$ by taking adjoints. And $(|A|+A)(|A|-A)=0$ gives $E(|A|-A)=|A|-A$. Hence,
$$
2EA=E(|A|+A)-E(|A|-A) = A-|A| \\
|A| = (I-2E)A \\
2E|A| = E(|A|+A)+E(|A|-A)=|A|-A \\
A = (I-2E)|A|.
$$
These two equations are consistent because $(I-2E)^2=I-4E+4E=I$ establishes $I-2E$ as its own inverse. Taking adjoints of the above equations shows that $E$ commutes with $A$ and with $|A|$, which is useful in what follows. Now the operators $P$ and $N$ may be written as
$$
P=\frac{1}{2}(|A|+A)=\frac{1}{2}(|A|+(I-2E)|A|)=(I-E)|A|, \\
N=\frac{1}{2}(|A|-A)=\frac{1}{2}(|A|-(I-2E)|A|)=E|A|
$$
Because $E$ commutes with $A$, then $E$ must also commute with $A^2$ and, hence, also with $|A|=(A^2)^{1/2}$. By the result of the first paragraph, $P=(I-E)|A|$ and $N=E|A|$ are positive.
First, you need $H$ to be a complex Hilbert space for this to be true even for bounded operators; it is false for the real Hilbert space $\mathbb{R}^2$, as witnessed by $A = \left( \begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\right)$.
But even for complex Hilbert spaces, your claim is in general not true. Most glaringly, you have no hypotheses that force $A$ to be closed. But it's not even true for closed operators. Take $H = L^2([0,1])$ and consider the operator $Af=-f''$ defined on the domain $D(A) = H^2_0([0,1])$ (i.e. the completion of $C^\infty_c((0,1))$ in the $H^2$ Sobolev norm). This is a positive closed operator, but it is not self-adjoint, since for instance the constant function 1 is in the domain of $A^*$.
(Indeed, you can get two different self-adjoint extensions of $A$ by allowing either Dirichlet or Neumann boundary conditions. The problem with the domain $H^2_0$ is that it's imposing both boundary conditions, which is too restrictive.)
For a positive result, however (no pun intended), it is true that every densely defined positive unbounded operator on a complex Hilbert space has at least one self-adjoint extension. This is the famous Friedrichs extension theorem.
Best Answer
The norm is at least $|2+3i|$ by definition, so it is not unitary which implies norm $1$. It can't be selfadjoint, because the eigenvalues of selfadjoint operators must be real. It could be normal though, because $A^*$ being the scalar times the identity is suitable.