Spectral Measures: Square Root Isometric Equality
Given a Hilbert space $\mathcal{H}$.
Consider a closed operator:
$$A:\mathcal{D}(A)\to\mathcal{H}:\quad A=A^{**}$$
Denote for shorthand:
$$H:=A^*A:\quad H=H^*$$
Regard elements:
$$\varphi\in\mathcal{D}(H)\subseteq\mathcal{D}(A)\cap\mathcal{D}(\sqrt{H})$$
A calculation gives:
$$\|\sqrt{H}\varphi\|^2=\langle\sqrt{H}\varphi,\sqrt{H}\varphi\rangle=\langle H\varphi,\varphi\rangle=\langle A\varphi,A\varphi\rangle=\|A\varphi\|^2$$

But for both it is a core:
  $$\overline{A_{\mathcal{D}(H)}}=A\quad\overline{\sqrt{H}_{\mathcal{D}(H)}}=\sqrt{H}$$

How can I prove the latter?
Partial Isometry
By the check above:
$$\varphi\in\mathcal{D}(A)=\mathcal{D}(|A|):\quad\||A|\varphi\|=\|A\varphi\|\quad$$
Construct the isometry:
$$U_0:\mathcal{R}(|A|)\to\mathcal{R}(A):\quad U_0(|A|\varphi):=A\varphi$$
Extend this uniformly:
$$U_E:\overline{\mathcal{R}(|A|)}\to\overline{\mathcal{R}(A)}:\quad U_E(\varphi):=\lim_nU_0(\varphi_n)$$
Lift to partial isometry:
$$U:\overline{\mathcal{R}(|A|)}\oplus\mathcal{R}(|A|)^\perp\to\mathcal{H}:\quad U(\varphi+\varphi^\perp):=U_E(\varphi)$$
The polar decomposition.
 A: Let $S$ be any densely-defined selfadjint linear operator on a Hilbert Space with spectral measure $E$. The range of $E[-\lambda,\lambda]$ is in the domain of every positive power of $S$. In fact,
$$
                    \mathscr{C}=\bigcap_{n=1}^{\infty}\mathcal{D}(S^{n})
$$
is a core for $S$. This is because the range of $E[-\lambda,\lambda]$ is in $\mathscr{C}$, and
$x\in\mathcal{D}(S)$ implies
$$
             \lim_{\lambda\uparrow\infty}E[-\lambda,\lambda]x = x,\\
             \lim_{\lambda\uparrow\infty}SE[-\lambda,\lambda]x = Sx.
$$
The last limit holds because of the Spectral Theorem characterization of the domain of $S$ as the set of $x$ for which $\int_{\mathbb{R}}\lambda^{2}\,d\|E(\lambda)x\|^{2} < \infty$.
In your case: Based on the phrasing of your comment it appears that you know $\mathcal{D}(A^{\star}A)$ is a core for $A$, and you want to know if $\mathcal{D}(A^{\star}A)$ is a core for $|A|$. So I'll only address the statement that $\mathcal{D}(A^{\star}A)$ is a core for $|A|$.
If $A$ is as you state, then $|A|=(A^{\star}A)^{1/2}$ has $\mathcal{D}(|A|^{2})=\mathcal{D}(A^{\star}A)$ as a core, based on the analysis for the case where $S=|A|$.
