How can I prove that a $2\times2$ matrix $A$ is area-preserving iff $\det(A)=1$ or $\det(A)=-1$?
A cute argument (in any dimension $n$) is to recall that the signed volume of a parallelepiped spanned by $v_1, \ldots, v_n$ is given by the determinant of the matrix
$$V = \left[ v_1, \ldots, v_n\right],$$
and that a matrix $A$ transforms the parallelepiped simply by $V \to AV$, i.e. by acting on the columns of $V$. Thus $A$ preserves area when
$$|\det V | = |\det(AV)| = |\det A||\det V|.$$
[This was going to be just a comment, but it got too large]
The answers of @Phira and @user7530 are both correct.
In fact, it suffices to see that $A$ preserves the areas of parallelograms. For an arbitrary region $E$, I may approximate $E$ by small rectangles (parallelograms), each centered at a specific vector $\mathbf{v}_i$ (think of integration to find area of a region in $\mathbb{R}^2$). Now say each small rectangle has area $a_i$. Then if $A$ is parallelogram-preserving, applying $A$ to the small rectangle centered at $\mathbb{v}_i$ will result in a parallelogram of size $a_i$ centered at $A\mathbb{v}_i$.
There is the small matter of showing that rectangles do not get mapped so that their images overlap, but that is easy to rule out because if true, then the area of a larger parallelogram containing the two would necessarily map to a parallelogram of less area.
Hope this helps!
The answer to this question has two parts:
The difficult part: A linear map $A$ multiplies the area of all shapes with the same factor $\chi_A$. To prove this one has to set up a clear notion of “area”. Then one proves that this “area” has the expected properties (finite additivity, various invariances, etc.) by means of an approximation argument, as hinted at in Shaun Alt’s answer.
The easy part: The factor $\chi_A$ described in 1. is given by $\chi_A=|\det(A)|$. This can be proven, e.g., in the following way: It is a theorem of linear algebra that any $2\times2$-matrix $A$ is the product of matrices of the following types:
$$\left[\matrix{\lambda & 0\cr 0& 1\cr}\right],\ \left[\matrix{1 & 0\cr 0& \mu\cr}\right],\ \left[\matrix{1 & 1\cr 0& 1\cr}\right],\ \left[\matrix{1 & 0\cr 1& 1\cr}\right].$$
For each of these types $T$ one can easily verify that $\chi_T=|\det(T)|$, and as $\chi_{\cdot}$ as well as $|\det(\cdot)|$ is multiplicative the general statement follows.