You are right to be hesitant. Having the same spectrum is a necessary, but not sufficient condition for similarity.
To show these two aren't equal, let be the first matrix, and be the second. Suppose for some invertible S.
Then we can adjust both sides by a scalar matrix and preserve similarity, i.e. for some scalar alpha, .
Then in particular, no matter how we shift the diagonal the ranks must stay the same on both sides of the equation since similarity preserves rank.
But if we take , obviously has rank 2 while has rank 1, thus a contradiction is reached and these two matrices are not similar.
If you are curious where this type of argument comes from, look up the Jordan Canonical Form. It is essentially the "best" representative from each similarity equivalence class, in that its structure makes obvious eigenvalue and eigenvector degeneracies and such rank arguments are essential to understanding that structure.