We are assuming that the matrices are diagonalizable. So, instead, you should show that they are similar to the same diagonal matrix $D,$ and conclude from there.
I would say that your counter-example is correct under your understanding of the terminology, but under what I believe to be a more common interpretation, you're incorrect.
The issue is that if someone tells me two matrices have the same eigenvalues, I assume that they mean counting multiplicity, so that if one matrix has eigenvalues $1,2,2$ then the one with the same eigenvalues must have eigenvalues $1,2,2$, not $1,1,2$.
Under this understanding, it is indeed true that two diagonalizable matrices with the same eigenvalues are similar. Write them as:
$$A_1 = P_1 D_1 P_1^{-1},$$
$$A_2 = P_2 D_2 P_2^{-1}.$$
If the diagonal elements on the two diagonal matrices are in the same order, then we can see similarity as
$$A_1 = P_1 P_2^{-1} D_2 P_2 P_1^{-1} =
P_1 P_2^{-1} D_2 (P_1 P_2^{-1})^{-1}$$.
If the diagonal elements happen to be in different orders, you can just use permutation matrices to swap the order. For example, applying a matrix permuting rows $i,j$ on both sides (this matrix is its own inverse) will swap the $(ii),(jj)$ elements of the diagonal matrix.
Best Answer
$B = P^{-1}AP \ \Longleftrightarrow \ PBP^{-1} = A$. If $Av = \lambda v$, then $PBP^{-1}v = \lambda v \ \Longrightarrow \ BP^{-1}v = \lambda P^{-1}v$. So, if $v$ is an eigenvector of $A$, with eigenvalue $\lambda$, then $P^{-1}v$ is an eigenvector of $B$ with the same eigenvalue. So, every eigenvalue of $A$ is an eigenvalue of $B$ and since you can interchange the roles of $A$ and $B$ in the previous calculations, every eigenvalue of $B$ is an eigenvalue of $A$ too. Hence, $A$ and $B$ have the same eigenvalues.
Geometrically, in fact, also $v$ and $P^{-1}v$ are the same vector, written in different coordinate systems. Geometrically, $A$ and $B$ are matrices associated with the same endomorphism. So, they have the same eigenvalues and geometric multiplicities.