As a disclaimer, I know the answer to this question; I'm sharing it here because I think others may enjoy tackling it.
(Notation: for $X\subseteq\Bbb R^2$, $~\mathbf1_X$ is the indicator function of $X$, defined by $p\mapsto\begin{cases}1,&p\in X\\0,&p\notin X\end{cases}$.)
Suppose $X_1,\dots,X_j$ and $Y_1,\dots,Y_k$ are subsets of $\Bbb R^2$ homeomorphic to $\Bbb R^2$. Suppose also that $\sum_{i=1}^j\mathbf1_{X_i}=\sum_{i=1}^k\mathbf1_{Y_i}$. Must $j=k$?
Said another way, suppose $X_1,\dots,X_j$ are subsets of $\Bbb R^2$ homeomorphic to $\Bbb R^2$, and $n_1,\dots,n_j$ are integers (or even reals). If $\sum_{i=1}^jn_i\mathbf1_{X_i}=0$, must $\sum_{i=1}^jn_i=0$?
Here's a counterexample. This was as simple as I could make it; I don't know if there's a simpler solution.
In the first image, you have two open sets (red and black), so $j=2$; in the second, you have four (red, black, blue, and green), so $k=4$. Their sums are identical, including where the overlaps are, but $j\ne k$.
Note, by the way, that you cannot do this with only finitely many intersections. For example, look at how the frame crops the first image. It cuts the red set into three pieces, so in the finite part of the figure that fits into the frame, both images have four sets. It's only in the infinite that it becomes only two sets.
My original counterexample was a lot more complicated; this went through some rounds of revision.