I played around with SAP's buckling feature and convinced myself that it doesn't handle cases like yours correctly. Your program also might not.
Then again, it's late on a Friday night, LOL, and I've been known to be wrong before.
I tried this problem: Pinned-pinned column, internally subdivided. Concentric load = 1 kip. Use eigenvalue analysis to get the multiplier. Multiply this by 1 kip and it matched the Euler buckling load perfectly.
Then I changed the problem to have a concentric load = 1 kip and a small moment. Remember the multiplier should be multiplying ALL loads. Run the buckling analysis and the answer is the same.
Then I changed the problem to have a concentric load = 1 kip and a very large moment--exact same result, obviously bogus. With the moment I put on there, a tiny P should cause huuuuuge displacements.
These buckling analyses work by solving the eigenvalue problem using the elastic stiffness matrix and the geometric stiffness matrix:
[Ke + lambda Kg]{phi}={0}
Solve for your load multiplier, lambda. For framed structures, Kg is filled with a bunch of constants * P/L. Shells should be similar, although more complicated.
Depending on your problem, the algorithm might do the same thing. If your eccentric load doesn't cause Kg's P's to be different, then the algorithm will not take them into account. I guess it depends on the problem--some eccentricities might change the P's.