Let $C\subset \omega \bigwedge A\bigcap B = \emptyset \bigwedge A\bigcup B = C$.
Let $\{x_i\}$ be a sequence of nonnegative reals.
Suppose $C$ is infinite and $\sum_{i\in C} x_i$ converges. (Since it converges absolutely, it makes sense to define its sum in this way)
Then $\sum_{i\in C} x_i = \sum_{i\in B} x_i + \sum_{i\in A} x_i$, when $A,B$ are infinite?
The sum of a countable family of non-negative numbers is the supremum of the set of sums of finite subfamilies. This definition does not use any order on the (index set of) the family. Since a finite subset of $C$ splits into a finite subset of $A$ and a finite subset of $B$, and conversely any two such finite subsets unite to a finite subset of $C$, you can easily prove $\sum_{i\in C} x_i = \sum_{i\in B} x_i + \sum_{i\in A} x_i$.
Your theorem is true if the sum converges absolutely.
If not, $x_n = \frac{(-1)^n}{n}$ is a counterexample. Observe that $$ \sum_{n=1}^\infty x_n$$ converges (since $x_n$ has alternating signs, and $\lim_{x\to\infty} x_n = 0$), but both of the sums $$ \begin{eqnarray} \sum_{n=0}^\infty x_{2n} &\text{ and }& \sum_{n=0}^\infty x_{2n+1} \end{eqnarray} $$ diverge.
Note, btw, that the notation $\sum_{i\in C} x_i$ already implies that the sum is invariant under at least some reorderings, since it doesn't specify any particular ordering. How trivial or non-trivial the proof of your theorem is depends on the precise definition you use for $\sum_{i\in C} x_i$.
EDIT: Only after posting the answer below I've noticed the question is only about non-negative series. I'll leave the answer here, since I think it might be useful anyway. (It is about a slightly more general question.)
We can even dispose the countability condition - we can have arbitrary index set.
Standard definition of absolute convergence of $\sum_{i\in I} x_i$, where each $x_i$ are real numbers, is that:
$S$ is the sum of this series if and only if for each $\varepsilon>0$ there exists a finite subset $F\subseteq I$ such that for each finite set $G\supset F$, $G\subseteq C$, we have $\left|\sum_{i\in G}x_i-S\right|<\varepsilon$.
This is the same definition as the one given in answers to this question, but I rewrote it in a way which avoids mentioning nets.
You can check that if $I=\mathbb N$ this is the same as saying that $\sum x_i$ converges absolutely to $S$. (For details see below.)
In fact we could use the same definition with $x_i\in X$ where $X$ is arbitrary linear normed space. It can be shown that if $X$ is a Banach space, then for sequences this is equivalent to uncoditional convergence, see e.g. Heil, A basis theory primer, p.93-94 or Theorem 14.9 in Pete L. Clark's Honors Calculus notes. It is not true that this is the same as absolute convergence, but in finitely dimensional Banach spaces this two notions coincide. Unconditional and absolute convergence are also discussed in this question.
Now if we work with the above definition and if $C=A\cup B$, where $A$, $B$ are disjoint and if we have $$S_1=\sum_{i\in A} x_i \qquad S_2=\sum_{i\in B} x_i$$ then for each $\varepsilon>0$ there is a finite subset $F_1\subseteq A$ such that for every finite $G$ fulfilling $F_1\subseteq G\subseteq A$ we have $$\left|\sum_{i\in G} x_i-S_1\right|<\varepsilon/2$$ and we also have $F_2$ such that for finite $G$ fulfilling $F_2\subseteq G\subseteq B$ we have $$\left|\sum_{i\in G} x_i-S_2\right|<\varepsilon/2.$$
Together we get (using triangle inequality) that for each finite set $G\subseteq C$ such that $G\supseteq F_1\cup F_2$ $$\left|\sum_{i\in G} x_i - (S_1+S_2)\right|<\varepsilon.$$
So we have shown that if $\sum_{i\in A} x_i=S_1$ exists and $\sum_{i\in B}=S_2$ exists, then $\sum_{i\in C}=S$ exists and it is equal to the sum $S=S_1+S_2$.
To see that if $\sum_{i\in C} x_i$ exists (in the sense of the above definition) then the sum $\sum_{i\in A} x_i$ exists we may use Cauchy criterion. The following is the formulation from Dixmier's General Topology:
9.1.6. Theorem (Cauchy's Criterion). Let $E$ be a Banach space, $(x_i)_{i\in I}$ a family of elements of $E$. The following conditions are equivalent:
(i) the family $(x_i)_{i\in l}$ is summable;
(ii) for every $\varepsilon > 0$, there exists a finite subset $J_0$ of $I$ such that, for every finite subset $K$ of $I$ disjoint from $J_0$, one has $\|\sum_{i\in K} x_i\| < \varepsilon$.
(By summable family the author means that $\sum_{i\in I} x_i$ exists in the sense of the above definition.)
Again, for the case of real numbers you can find this result in Pete L. Clark's notes Honors Calculus as Theorem 14.5 (Cauchy Criteria for Unordered Summation). (Including the proof.)
It is relatively easy to see that if $A\subseteq C$ and $\sum_{i\in C} x_i$ fulfills Cauchy condition, then so does $\sum_{i\in A} x_i$.
Now back to the case $I=\mathbb N$.
Suppose that a series $\sum_{i=1}^\infty x_i$ is absolutely convergent to $S$. Then we also have $\sum_{i=1}^\infty |x_i|<\infty$. This implies that, for any given $\varepsilon>0$, there is $n_0$ such that $$\left|\sum_{i=n_0+1}^{n_0+k} x_i\right|\le\sum_{i=n_0+1}^{n_0+k} |x_i|<\varepsilon/2$$ holds for each $k$. We can choose $n_0$ such that, in addition to the above condition, $\left|\sum_{i=1}^{n_0}-S\right|<\varepsilon/2$. So we can choose $F=\{1,2,\dots,n_0\}$ and we get that for any $\{1,2,\dots,n_0\}\subseteq G\subseteq\{1,2,\dots,n_0+k\}$ we also have $$\left|\sum_{i\in G\setminus F} x_i\right| \le \sum_{i\in G\setminus F} |x_i| \le \sum_{i=n_0+1}^{n_0+k} |x_i| < \varepsilon/2$$ which together with $$\left|\sum_{i\in F} x_i-S\right| < \varepsilon/2$$ implies $$\left|\sum_{i\in G} x_i-S\right| < \varepsilon.$$
On the other hand, if $\sum_{i\in\mathbb N} x_i$ fulfills the above condition then each reordering $\sum_{i=1}^\infty x_{\pi(i)}$ is convergent. (For any given $\varepsilon>0$ we choose $n_0$ large enough so that $\{x_{\pi(1)},\dots,x_{\pi(n_0)}\}\supseteq F$.) This implies that $\sum_{i=1}^\infty x_i$ is absolutely convergent.
