How simple can the PRP used in the Even-Mansour scheme be?
The permutation used for Even-Mansour needs to approximate a random permutation. This is the model in which all of the $r$-round Even-Mansour proofs are done. If you have a permutation that is distinguishable from random in a small number of queries, those proofs become null and void—though in some cases the cipher may still be secure.

Now, it is perfectly reasonable to wonder about the exact limits of this approach like you did, and in fact people have: this is a popular cipher design approach otherwise known as "key-alternating cipher". The AES itself is probably the most famous instance of it. You can model the AES as 
$$
P'(\dotsc P(P(m \oplus k_0) \oplus k_1) \oplus k_2) \dotsc) \oplus k_{10},
$$
where $$P(x) =\mathsf{MixColumns}(\mathsf{ShiftRows}(\mathsf{SubBytes}(x)))$$ and $$P'(x) = \mathsf{ShiftRows}(\mathsf{SubBytes}(x)),$$ and $k_i$ are the keys produced by the key schedule.

While the AES is not provably secure, since each of those permutations is way too simple to behave as a random permutation, heuristically this works out against known attacks given enough rounds. But in this case, the Even-Mansour results are of little use (besides confirming that the high-level design approach is sound), and concrete analysis against differential, linear, etc cryptanalysis is much more valuable.