Consider first the expansion of the electrostatic potential $\Phi$ in a stationary situation $$\Phi(\vec{r}) = \frac{Q}{4 \pi \epsilon_0 r} + \frac{\vec{d}\cdot \vec{r}}{4\pi \epsilon_0 r^3} + \mathcal{O}(r^{-3})$$ Now let us shift our origin by $\vec{a}, \vec{R} = \vec{r} - \vec{a}$, where $|\vec{a}| \ll r$. The monopole term then gets expanded as $$\frac{Q}{4 \pi \epsilon_0 r} = \frac{Q}{4 \pi \epsilon_0 |\vec{R} + \vec{a}|} = \frac{Q}{4 \pi \epsilon_0 R} - \frac{Q(\vec{a}\cdot \vec{R})}{4 \pi \epsilon_0 R^3} + \mathcal{O}(R^{-3})$$ The dipole term is simply $$\frac{\vec{d}\cdot \vec{r}}{4\pi \epsilon_0 r^3} = \frac{\vec{d}\cdot \vec{R}}{4\pi \epsilon_0 R^3} + \mathcal{O}(R^{-3})$$ Now let's take a look at the whole potential $$\Phi(\vec{r}) = \Phi(\vec{R}) = \frac{Q}{4 \pi \epsilon_0 R} + \frac{(\vec{d}-Q\vec{a})\cdot \vec{R}}{4\pi \epsilon_0 R^3} + \mathcal{O}(R^{-3})$$ Note that this is (within the approximation) just a different description of the same potential as above. In other words, if you take a look at the same physical point, you will get the same value of $\Phi$ independent of whether you are in the $\vec{r}$ coordinates or the $\vec{R}$ coordinates.
Now let us denote the dipole defined with respect to $\vec{R}$ as $\vec{D}$ and compute $$\vec{D} = \int \rho \vec{R} d V = \int \rho \vec{r} d V - \int \rho \vec{a} d V = \vec{d} - Q\vec{a}$$ We then see that our shifted potential in $\vec{R}$ coordinates can be written as $$\Phi(\vec{R}) = \frac{Q}{4 \pi \epsilon_0 R} + \frac{\vec{D}\cdot \vec{R}}{4\pi \epsilon_0 R^3} + \mathcal{O}(R^{-3})$$ which is exactly the multipole expansion that you would get if you started from the $\vec{R}$ coordinates in the first place.
I.e., multipole expansions are covariant with respect to coordinate shifts. It is possible to show that this applies to all orders with more and more terms trickling down from lower orders to higher ones, and you can even show covariance of the multipole expansion with respect to the whole <PERSON> group (with small translations).
You are now probably asking what is happening with your radiation formula then. 649
The trick is that the formulas you give will apply only in inertial frames. In particular, $\vec{a}$ will typically be a constant shift. However, your radiation formula applies to oscillation magnitudes to which a constant $\vec{a}$ will have subleading or outright vanishing contributions.
Consider the dipole moment. We have $\vec{D} = \vec{d} - Q\vec{a}$. Then you see that since $\dot{Q} = \dot{a} = 0$, we have $\dot{D} = \dot{d}$ and there is no extra term arising in the radiation formula from the shift.
As for the dipole magnetic moment, we have $$\vec{\mu}' = \vec{\mu} + \frac{1}{c}\vec{a} \times \int \vec{j}\, dV$$ It is a slightly more complicated argument why the extra term will be subleading. 499