No, the method in your title cannot be used in your case, if you want the exact answer. Your physics colleagues are correct. Integration (as explained by probably-someone) gives the exact answer.
Replacing the distributed mass by a point mass at the centre of mass works for a spherically symmetric (3D) distribution, but not (in general) for any other distribution. For a spherically symmetric distribution this is called Newton's Shell Theorem. By extension this applies also to the force between two spherically symmetric masses. It does not work for a 3D ellipsoid, unless the ellipsoid is spherical.
As JMLCarter suggests, your method gives an approximate answer, and this approximation becomes better as the separation becomes much bigger than the dimensions of the ellipse. But this is trivial, I think, because in the limit of large separations all finite objects
look like point particles; any departure from spherical symmetry becomes insignificant.
The Shell Theorem does not work in 2D, so it does not work for an elliptical disk or ring, not even if the disk or ring is circular. In order for it to work for circular distributions of mass, the force law would have to be prortional to $1/r$ instead of $1/r^2$. See Newton's Gravitation Force between two bodies and Gravitational field of thin 2D ring - numerical simulation. So it will work for the force between a line mass and a cylindrical shell or solid, provided that their axes are parallel and both are infinitely long - ie much longer than the distance between them.
