mjhjmtu

What is wrong with this proof:
Suppose R and T are similar operators and R has eigenvalue 2 for some eigenvector $v$.
By property of similar matrices:

$R=STS^-1$

Therefore:

$Rv = STS^-1v$

$2v = STS^-1v$

$2S^-1v = TS^-1v$

$2S^-1Sv = Tv$

$2Iv = Tv$

$Tv = 2v$

Thus $v$ is also an eigenvector of $T$ with eigenvalue 2.
Clearly this proof is incorrect, but where does it go wrong?

When you went from
$$
2S^-1v=TS^-1v
$$
to
$$
2S^-1Sv=Tv
$$
you completely ignored the rules of matrix multiplcation; you cannot arbitrarily commute matrices in a product! It should have read as follows:
$$
2S^-1vS=TS^-1vS
$$
Also not that it is not even clear that this is well defined as written, since $v$ is probably a column vector.

Since the error in our OP Justin Sanders' argument has been thoroughly vetted in other answers, here I will directly address the title question.

Similar matrices do have the same eigenvalues, to wit:

if

$B = PAP^-1, tag 1$

then

$B - lambda I = PAP^-1 - lambda I = PAP^-1 - lambda PIP^-1 = P(A - lambda I)P^-1, tag 2$

whence

$det(B - lambda I) = det(P(A - lambda I)P^-1) = det(P) det(A - lambda I) det (P^-1) = det(A - lambda I), tag 3$

since $det(P^-1) = (det(P))^-1; tag 4$

thus $A$ and $B$, having the same characteristic polynomials, also share the roots of these polynomials, i.e., their eigenvalues.

However, similar matrices do not in general share eigenvectors; if

$B vec v = lambda vec v, tag 5$

then

$PAP^-1 vec v = lambda vec v, tag 6$

or

$AP^-1 vec v = lambda P^-1 vec v, tag 7$

that is, $P^-1 vec v$ is an eigenvector of $A$ corresponding to $lambda$; indeed, if $lambda$ is a root of (3) of multiplicity one, then $P^-1 vec v$ is, up to a scale factor, the eigenvector of $A$ associated with $lambda$. Since we can in general choose $P$ so that $P^-1 vec v ne vec v$, the eigenvectors will not be shared 'twixt' $A$ and $B$.

So while the eigenvalues are similarity invariant, the eigenvectors transform according to $vec v mapsto P^-1 vec v$.

It's wrong when you jump from $2S^-1v=TS^-1v$ to $2S^-1Sv=Tv$.