Is it possible to find two infinite sets of non-negative integers, $A,B$, such that every non-negative integer can be written uniquely as a sum of two integers, one from $A$ and the other from $B$?
It is easy to do this if we let one set be finite. For example, take $A=\{0,1\}$ and let $B$ be the set of non-negative even integers. Can it be done with two infinite sets?
Remark: the assumptions imply that $A\cap B=\{0\}$. Indeed, the only way to write $0$ as the sum of two non-negative integers is $0=0+0$ so $0$ must be in both. But then, if we had a non-zero $n\in A\cap B$, we could then write $n=n+0=0+n$, contradicting the uniqueness of the decomposition.
Take the decomposition in base $2$ (or your favourite number), any $n \geq 0$ can be written in an unique way as $\sum_{k=0}^{m} a_k 2^k$. Now separate between odd and even powers, $$n = \sum_{k=0}^{\lfloor \frac{m}{2} \rfloor} a_{2k} 2^{2k} + \sum_{k=0}^{\lfloor \frac{m-1}{2} \rfloor} a_{2k+1} 2^{2k+1}$$
Take $A$ to be the integers that are the sum of even powers of $2$ (and $0$), and $B$ the ones of odd powers (and $0$), ie $$ A = \{ \sum_{k=0}^m a_k 2^{2k} \mid m \geq 0, (a_0,\ldots,a_m) \in \{0,1\}^{m+1} \} \cup \{0\}$$ $$ B = \{ \sum_{k=0}^m a_k 2^{2k+1} \mid m \geq 0, (a_0,\ldots,a_m) \in \{0,1\}^{m+1} \} \cup \{0\}$$
Then surely $A \cap B = \{ 0 \}$, any $n \geq 0$ can be written as an element of $A + B$ and this sum is unique because decomposition in base $2$ is.