Predicting color of precipitate
There are a few wrong assumptions in your question that I would like to clear up.


Yes, d-orbital splitting is part of the reason for colour. However, $\ce{[FeF6]^3-}$ displays a beautiful d-orbital splitting which could display colour from the energy difference — but doesn’t. There is more to colour than d-orbital splitting; the most important terms are spin-forbidden and Laporte rules.
You somehow got the impression that ionic solids have nothing to do with coordination complexes. Please remove this idea from your brain; it is wrong. Ionic solids are equally coordination complexes as present in solution only that the environment and sometimes the composition of the complex is different. E.g, $\ce{CuSO4(aq)}$ is essentially a solvated sulphate ion plus $\ce{[Cu(H2O)6]^2+}$. When precipitated or crystallised as ‘$\ce{CuSO4 . 5H2O}$’, what you actually got was $\ce{[Cu(H2O)4(\mu{-}SO4)2].H2O}$ — clearly a coordination compound.


To answer your question, it depends on the nature of the compound you are precipitating — the colour can derive from cation, anion or their combination. For example, aluminium salts will be colourless unless the anion dictates otherwise. Likewise for zinc salts. For a very large number of salts where the cation is responsible for the colour, these will be the same as in solution. But sometimes ($\ce{PbI2, Fe4[Fe(CN)6]3}$ and others) you just have to know because only this specific combination will give this colour.