Suppose that the work set is $[-\infty+\infty]$, we suppose that $$\sum_{n=0}^{+\infty}a_n<+\infty.$$ Now can we says that $$\sum_{n=0}^{+\infty}(a_n-b_n)=\sum_{n=0}^{+\infty}a_n-\sum_{n=0}^{+\infty} b_n\quad$$
In the $[-\infty,+\infty]$ each series(well defined) converges, therefore for me is yes. It's true?
If $\displaystyle\sum_{n=1}^\infty a_n = -\infty$, then the result need not hold. Consider the counterexample:
$$a_n = -1, b_n = -1.$$
If we assume further that $-\infty < \displaystyle\sum_{n=1}^\infty a_n$, then it does hold (in a reasonable sense) as I shall show now.
Let $S_N = \displaystyle\sum_{n=1}^N a_n - b_n,$ $A_N = \displaystyle\sum_{n=1}^N a_n,$ and $B_N = \displaystyle\sum_{n=1}^N b_n.$
You wish to answer whether the equality $$\lim_{N \to \infty}S_N = \lim_{N \to \infty}A_N - \lim_{N \to \infty}B_N$$ holds.
Clearly, we have $S_n = A_n - B_n$ for all $n \in \Bbb N$.
The convergence of $(A_n)$ (to a real number) is part of the hypothesis.
If $(B_n)$ converges to a real number, then it follows that your desired equality holds.
If $(B_n)$ converges to $-\infty$, then the RHS is $\infty$. Moreover, we can bound $A_n$ and thus, it is easy to show that $S_n \to \infty$ as well. Thus, the equality holds again.
The same consideration applies to the case $B_n \to \infty$.
Now, let us consider the case that $(B_n)$ diverges (that is, not converges). In this case, we have that the RHS is not defined.
We show that the LHS isn't defined either. That is, show that $(S_n)$ diverges as well.
Note that $$B_n = A_n - S_n \quad \forall n \in \Bbb N.$$ If $(S_n)$ were convergent, then $(A_n - S_n)$ would be as well and in turn, so would $B_n$. This gives us a contradiction and thus, we see that $(S_n)$ is divergent.
Thus, we conclude as follows:
Equality does hold in the following sense:
One side of the equality is defined if and only if the other is. Moreover, whenever both sides are defined, they are equal.