The sum of integers being a bijection

Here is a fairly large class of examples. Pick any subset $S$ of $\mathbb{N}$. Let $P$ be the set of non-negative integers such that the only $1$s in their binary expansion are at indices in $S$, and let $Q$ be the set of non-negative integers such that the only $1$s in their binary expansion are at indices in the complement of $S$. (Your examples are, respectively, $S = \mathbb{N}$ and $S = \mathbb{N} - \{ 0 \}$.) Similar constructions work for any base, and for slightly more general things than bases (e.g. factorial base). In terms of infinite series this is a consequence of the identity

$$\frac{1}{1 - x} = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)...$$

which expresses the uniqueness of binary expansion, and the choice of $S$ corresponds to a choice of grouping of terms on the RHS.

To comment on Qiaochu's answer, one can show that all such factorizations come from mixed radix representations (different bases, factorial base etc.). That is if $$\frac{1}{1-x}=P(x)Q(x)$$ then there must be a sequence $1=a_0\le a_1 \le a_2\le\cdots$ so that $a_i$ divides $a_{i+1}$ and disjoint subsets $A,B$ with $\mathbb N=A\cup B$, so that $$P(x)=\prod_{i\in A}\frac{1-x^{a_{i+1}}}{1-x^{a_i}},Q(x)=\prod_{i\in B}\frac{1-x^{a_{i+1}}}{1-x^{a_i}}.$$ The proof is simple, suppose $P(x)=1+x+\cdots +x^{a_1-1}+\cdots$ then $Q(x)=Q_1(x^{a_1})$ and $P(x)=\frac{x^{a_1}-1}{x-1}P_1(x)$. Then we proceed by induction.

If you accept that 0 is not a natural number, then there is a very simple answer to your question: take $P$ to be all numbers whose expansions base 4 contain only digits 0 and 1 and $Q$ to contain only digits 0 and 2. Then $P\cap Q=\{0\}$, which we have boldly excluded.

Also, both sets have the lowest possible asymptotic density of order $1/\sqrt n$, which is kinda nice.