This is a remark rather than an answer to your question. If you remove the word `simple' it is easy to find such pairs of finite groups. The first examples I learned were (I think) constructed by Atiyah. Each of the two finite groups has a normal subgroup that is cyclic of order three and quotient dihedral of order 8. In each case the dihedral group acts non-trivially on the $C_3$, but in one case the kernel of the action is $C_2\times C_2$ whereas in the other case the kernel of the action is $C_4$. The Lyndon-Hochschild-Serre spectral sequence gives a complete calculation of the integral cohomology ring, and it is just $H^*(C_3)^{C_2}\otimes H^*(D_8)$ in each case. Somehow cohomology cannot see that the centralizers of the $C_3$ subgroups are different (which is how one can see that the groups are not isomorphic).

I computed the integral cohomology rings of a family of $3$-groups of nilpotence class two in my thesis, and I observed that for each $n\geq 5$ there are a pair of groups of order $3^n$ with isomorphic integral cohomology rings. I also gave a more conceptual proof that there are pairs of groups of order $p^n$ for each odd $p$ and $n\geq 5$ that are not isomorphic but have isomorphic integral cohomology groups. My articles are '$3$-groups are not determined by their integral cohomology rings' JPAA Vol 103 (1995) 61-79 and '$p$-groups are not determined by their integral cohomology groups' Bull London Math Soc Vol 27 (1995) 585-589.

None of these arguments are any help for finite simple groups though.