No, the "integral" elements are not a subring. There are at least two ways to understand this.

In the case of $D$ being the integral Hamiltonian quaternions, or even the Hurwitz integers therein (adjoining $(1+i+j+k)/2$ to give a maximal subring), we can easily conjugate ourselves to another subring: using the model of $D$ inside two-by-two complex matrices spanned over $\mathbb R$ by $\pmatrix{1 & 0 \cr 0 & 1}$, $\pmatrix{i & 0 \cr 0 & -i}$, $\pmatrix{0 & 1 \cr -1 & 0}$, and $\pmatrix{0 & i \cr i & 0}$, conjugating by diagonal elements $\pmatrix{a+bi & 0 \cr 0 & a-bi}$ where $a\pm bi$ have some odd Gaussian prime factors *not* in common will move us out of the Hurwitz integers.

This is a direct manifestation of the point that subrings of $D$'s finitely-generated as $\mathbb Z$-modules are obtained by taking products of maximal compact subrings of all the $D\otimes_{\mathbb Q} \mathbb Q_p$ and intersecting with $D$. At primes $p$ where $D$ becomes the matrix algebra, there is by-far *not* a unique such maximal subring. Approximating this globally gives the kind of counterexample in the previous paragraph.