If - as stated in a comment by the OP - the area is enclosed by a perimeter that connects every point and doesn't intersect itself, then we might have a picture like this:
What we can do
is to draw ($\color{red}{red}$) lines from the origin $(0)$ to the vertices of
the polygon (six in our example) and sum up the areas of the triangles
$(0,1,2),(0,2,3),(0,3,4),(0,4,5),(0,5,6),(0,6,1)$ : see figure
on the left.
It is shown in the figure on the right how the area of just one of these triangles is calculated, using a determinant:
$$
\mbox{area}\,\Delta =
\frac{1}{2} \det\begin{bmatrix}(x_1-x_0) & (x_2-x_0)\\(y_1-y_0) & (y_2-y_0)\end{bmatrix}
$$
Note that, in general, the triangle areas thus calculated can be positive as well
as negative; and the latter is essential. Now continue for the vertex coordinates of the polygon,
in anti-clockwise order, to calculate the area of the whole 2-D object.
Closed perimeters in an image are likely to be the
product
of a contouring procedure.
Such contours or isolines can be clockwise or anti-clockwise oriented, corresponding respectively
with a negative or a positive enclosed area. It's wise not to destroy that information prematurely. As an example:
negative areas enable the existence of objects with holes in it.
In principle, the origin can be chosen at will. But for best results (numerically) the vertex centroid
$\,\sum_k(x_k,y_k)/N\,$ of the polygon may be a good choice.
It's easy to prove the OP's conjecture for one of these triangles.
One of the vertices is located at the origin, take $\,(x_0,y_0) = (0,0)$ . Therefore the triangle is defined by two column vectors, that can be summarized into a matrix. And the area of
that triangle is half the determinant of that matrix, as we have seen. Symbolically:
$$
\Delta = \begin{bmatrix} x_1 & x_2 \\ y_1 & y_2 \end{bmatrix}
$$
Now let the transformation be given by:
$$
T = \begin{bmatrix} a & b \\ c & d \end{bmatrix}
$$
Then the deformed triangle is represented by:
$$
\Delta' = \begin{bmatrix} x'_1 & x'_2 \\ y'_1 & y'_2 \end{bmatrix} =
\begin{bmatrix} a & b \\ c & d \end{bmatrix}
\begin{bmatrix} x_1 & x_2 \\ y_1 & y_2 \end{bmatrix} =
\begin{bmatrix} ax_1+by_1 & ax_2+by_2\\
cx_1+dy_1 & cx_1+dy_2\end{bmatrix}
$$
The deformed area, for one triangle, is:
$$
\frac{1}{2} \det\begin{bmatrix} x'_1 & x'_2 \\ y'_1 & y'_2 \end{bmatrix} =
\frac{1}{2} \det\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}
\begin{bmatrix} x_1 & x_2 \\ y_1 & y_2 \end{bmatrix}\right) =
\det(T)\left(\mbox{area}\,\Delta\right)
$$
And for all triangle areas summed:
$$
\sum_\Delta \left(\mbox{area}\,\Delta\right) \det(T) =
\det(T) \sum_\Delta \left(\mbox{area}\,\Delta\right) =
\mbox{Area of object} \cdot \det(T)
$$
Almost as conjectured. It is not guaranteed, though, that $\det(T)$ is positive and
that the absolute value of the determinant can be taken, unless such is specified explicitly for the transformation $T$.
For example, if we have mirroring in the y-axis:
$$
T = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \quad \Longrightarrow \quad \det(T) = -1
$$
If our perimeter has been counter-clockwise then it becomes clockwise, and the positive area
of the original is transformed into a negative area. We should put $\;\left|\det(T)\right|\;$ instead
of $\;\det(T)\;$ only if it is decided that the sign of the area of an object is not relevant.
