If $(X,\mathcal{S})$ is standard Borel space and $\mu$ a continuous measure on $(X,\mathcal{S})$ then there is a Borel isomorphism $F:X\to [0,1]$ that sends $\mu$ to Lebesgue measure on $[0,1]$. (See Kechris 17.41.) Since the isomorphism preserves measure this shows that any measurable subset of $[0,1]$ has measurable preimage under $F$. In other words, if all sets of reals are measurable then so are all subsets of $X$. So one does not have to look at the specifics of Solovay's model, nor at Haar measure for particular groups, as long as one restricts attention to Polish measure spaces.
It should be added that these arguments only depend on DC which does hold in Solovay's model. Without DC I expect very strange behaviour is possible.
