I found the image below illustrating the transformations by several matrices with given eigenvectors, eigenvalues and determinants. For matrix in the second row ($\lambda_1=\lambda_2=1$ and $det(A)=1$), the author wrote:
The eigenvalue $\lambda_1 = 1 = \lambda_2$ is repeated and the eigenvectors are collinear (drawn here for emphasis in two opposite directions). This indicates that the mapping acts only along one direction (the horizontal axis).
(I'm only mentioning the second row in this question.) However, I couldn't relate "the mapping acts only along one direction (the horizontal axis)" with the resulted visual transformation on the right. More specifically, given the two eigenvectors and eigenvalues as such and without looking at the transformed image on the right, I don't know for sure to how the square on the left should transform into. That is to say, what the shearing angle of the transformed parallelogram (the angle between the horizontal sides of the parallelogram on the right with the $x$-axis) and shearing direction (counter-clockwise or clockwise) are.
Can we identify the transformation (i.e. characterizing the parallelogram on the right) by solely looking at the eigenvectors, eigenvalues and determinant without having to reconstruct the transformation matrix from them?
Not a fan of that second sentence in your quote, so lets try to talk it out.
Probably the best idea here is to try to work out some matrices that have the quality that you want. That's hard to do when you're just learning the stuff, but if you know the ropes, you might come up with something like
$$A = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}. $$
What's going on? Well, if you solve the linear equation $(A-I)x = 0$ to find the eigenvectors, you realise that, so long as $a$ is non zero, there's only one family of them, ie $(1, 0)^T$ ie we're mapping the x axis to the x axis. Note that there is only one dimension of the solution, there's essentially only one eigenvector.
[as an aside, if we'd picked a different repeated eigenvalue, then all of what I would be saying would be the same, except with 'and dilated after that'. Also, if $a$ is $0$, the transform is hte identity, and I assume you understand how that works.]
So, what is happening in the $y$ direction? Well, the transform is acting like
$$ \begin{pmatrix} x \\ y \end{pmatrix} \mapsto \begin{pmatrix} x + ay \\ y \end{pmatrix} $$
In words, we might say that horizontal lines of height $y$ are mapped back onto themselves. But you can see, I picked $a$ arbitrarily, and it doesn't change the eigenvector or the eigenvalues, but it does change the skew.
So, the plain answer to your questions is "no we can't".
The complicated answer to your questions is "another year of linear algebra, generalised eigenvectors, minimal polynomials, and Jordan normal form". For most students, I would recommend taking the "no".