Diffie-Hellman when $p$ and $g$ are pre-agreed and hidden from Eve
For the moment assume $g$ is a secret (uniformly random) generator, but that $p$ may be known to the adversary. Then given only $g^a, g^b$, the Diffie-Hellman key $g^{ab}$ is information-theoretically uniform (up to small statistical error), i.e., it cannot even be found by brute force because the adversary does not have enough information to determine it. This is because every generator $h=g^c$ for all $c$ (coprime to the group order) is equally likely to be the "real" one, so the "real" key is equally likely to be $g^{ab/c^2}$. Such values essentially cover all group elements (up to rare degeneracies). So essentially, Alice and Bob have a true one-time pad value.

However, this is not really interesting, because Alice and Bob were able to set up a secret generator $g$ in the first place, which is just as good as (or better than) a one-time pad value.

If Alice and Bob use the same $g$ for two or more runs of the protocol, then there is only one "degree of freedom" in the derived keys. That is, knowing one key (and the public messages) information-theoretically determines all the others. So any hope that the keys would look uniform and independent would have to depend on a computational assumption.

This all assumed that $p$ was known. It's much harder to analyze what happens when $p$ is unknown, because revealing group elements reveals some information about $p$, but it's hard to say what.