This proof is different from the one in Denis Serre's book.
As usual, take $M^A$ and $M^B$ to be the $k[t]$-modules with underlying space $k^n$, where $t$ acts by $A$ and $B$ respectively. Then $A$ is similar to $B$ if and only if $M^A$ and $M^B$ are isomorphic as $k[t]$-modules. As $k[t]$ modules $M^A$ and $M^B$ are both generated by the coordinate vectors $e_1,\dotsc,e_n$, and given by relations (in matrix form)
$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(A-It)=0$
and
$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(B-It)=0$
In general, modules given by matrices of relations are isomorphic if and only if the relation matrices are equivalent.
Thus $A$ and $B$ are similar if and only if $A-tI$ and $B-tI$ are equivalent.