Suppose I have some vector $x = a + b = (a_0 + b_0, \dots, a_{n-1} + b_{n-1})$ of length $N$. Now, if only $x$ is known, what is the minimum amount of information that is needed to either solve or approximate what is $a-b$? Of course, if there is no any other information about $a$ and $b$ except their sum this cannot be solved.
To start, notice that the solutions of $a-b$ lie along the line $y = 2a - x$. To throw some idea, how much the number of solutions could be narrowed if I knew for example what is $\sum_{i=0}^{N-1} a_i - b_i$? The minimal information on $a$ and $b$ can be something else as well, if it's computable in linear time.
You are trying to solve for $2N$ variables. Your initial information about $a+b$ gives $N$ linear equations in these variables. If you add more linear equations, you will need at least $N$ more to have a chance of solving.
The one you propose $\sum_i a_i-b_i$, is one linear equation. Now you need $N-1$ more.
In the comments, you propose the $N$ linear equations $$C_K=\sum_{i=0}^K a_i-b_i$$ as $K$ varies from $0 $ to $N-1$. If you know all of these, then indeed you can recover what you want. In fact, you don't even need $a+b$ if you have all these. You have $$a-b=(C_0,C_1-C_0, C_2-C_1, \ldots, C_{N-1}-C_{N-2})$$