Short answer: Not really.
The answer is slightly different when talking about point charges vs distributons.
Given I have some charge distribution $\rho_{1}$ and some other charge distribution $\rho_{2}$ that produces a potential $V_{2}$
The potential energy between the 2 distributions [not the same thing as the total potential energy of the whole distribution] is given as the following:
$$\iiint V_{2} \rho_{1} dv $$
Aka building up distribution 1 in the presence of potential 2. This represents the potential energy between the 2 distributions.
Moving the distribution 1 to its location, as a whole, or as individual $\rho_{1} dv$ elements yield the same result. The potential energy between the distributions is invariant to the proccess, just the final state.
Now... this kind of goes out the window for point charges, sort of.
Point charges cannot be broken up into individual elements, so the question about multiple "dq" of a point charge doesn't really make sense. With that being said, a point charge has a single "dq" element, That is $Q\delta^3(\vec{r}-\vec{r}_{0}) dv$ Plugging it into our formula gives the same result, the potential energy between the distributions is invariant to how you arrive at the final distribution.
Edit:
This is also the same for total field energy , moving it as a whole requires you to build to distribution in the first place adding work in overcoming the electrostatic forces of itself. Then overcoming the forces from the other charges. Or alternatively, moving the charge piece by piece taking into account that the amount of work required to move each charge increases as you now need to overcome the forces of the placed charge.
Both of these methods yield the same total field energy
[May comment on the false infinite energy of a point charge later]
