There are no such two non-isomorphic groups. As is pointed out in Neil Strickland's answer, up to some exceptions, a finite simple group is determined by its order. For these exceptions it is known that the smallest character degree larger than 1 is different (For the infinite series $B_n(q)=O_{2n+1}(q)$ vs $C_n(q)=PSp_{2n}(q)$ for $q$ odd this follows from the results in Landazuri and Seitz, On the minimal degrees of projective representations of the finite Chevalley groups, J. (J. Algebra 32 (1974), 418–443, MR0360852 (50 #13299)). This means that a finite simple group is determined by its character degrees (with mulitplicities), and led Huppert (Some simple groups which are determined by the set of their character degrees I, Illinois J. Math. 44, 4 (2000), 828-842, MR1804317 (2001k:20009)) to the conjecture:
Conjecture: If two finite groups $G$ and $H$ have the same set of character degrees (without counting multiplicities) and $G$ is nonabelian simple, then $H\cong G\times A$ for some abelian $A$.
This has been verified for some simple groups, but is still open to the best of my knowledge. Needless to say that all this depends heavily on the classification.
