Solution:
First solution. It suffices to consider the case when $P$ and $Q\not\equiv 0$ are relatively prime polynomials and the leading coefficient of $Q$ equals $1$. We have
$$x(x+2)(P(x) Q(x+1)-Q(x) P(x+1))=Q(x) Q(x+1)$$
for infinitely many $x$, i.e. for every $x$. Thus the polynomials $Q(x)$ and $Q(x+1)$ divide $x(x+2) Q(x+1)$ and $x(x+2) Q(x)$ respectively.
Therefore $S(x) Q(x)=x(x+2) Q(x+1)$ and $T(x) Q(x+1)=x(x+2) Q(x)$, where $S$ and $T$ are quadratic polynomials with leading coefficients $1$. Hence, $S(x) T(x)=x^{2}(x+2)^{2}$. There are three cases to be considered.

Case 1. $S(x)=T(x)=x(x+2)$. Then $Q(x+1)=Q(x)$, i.e. $Q \equiv 1$ and the condition of the problem shows that this is impossible.

Case 2. $S(x)=x^{2}$ and $T(x)=(x+2)^{2}$. Then $x Q(x)=(x+2) Q(x+1)$. Therefore $Q(1)=0$ and it follows by induction that $Q(n)=0$ for all $n \in \mathbb{N}$. Hence $Q \equiv 0$, a contradiction.

Case 3. $S(x)=(x+2)^{2}$ and $T(x)=x^{2}$. Then $(x+2) Q(x)=x Q(x+1)$ and therefore $x$ divides $Q(x)$ and $x+2$ divides $Q(x+1)$, i.e. $x+1$ divides $Q(x)$. It follows that $Q(x)=x(x+1) Q_{1}(x)$, where $Q_{1}$ has leading coefficient $1$ and $Q_{1}(x+1)=Q_{1}(x)$. We conclude that $Q_{1}(x)=1$ and $Q(x)=x(x+1)$. Now plugging $Q(x)$ in (1) gives
$$(x+2) P(x)-x P(x+1)=x+1$$
Setting $x=0$ and $x=-1$ we obtain $P(0)=\frac{1}{2}$ and $P(-1)=-\frac{1}{2}$. Therefore $P(x)=\frac{1}{2}+x+x(x+1) P_{1}(x)$, where $P_{1}$ is a polynomial. Now (2) implies that $P_{1}(x+1)=P_{1}(x)$ and therefore $P_{1}$ is a constant.
We conclude that if the polynomials $P$ and $Q$ are relatively prime and $a_{0}=1$, then $Q(x)=x(x+1)$ and $P(x)=\frac{1}{2}+x+c x(x+1)$.
Therefore the answer is
$$Q(x)=x(x+1) R(x) \text{ and } P(x)=\left(\frac{1}{2}+x+c x(x+1)\right) R(x)$$
where $R$ is an arbitrary nonzero polynomial and $c$ is a constant.

Second solution. The given identity can be written as
$$\frac{P(x)}{Q(x)}-\frac{1}{2}\left(\frac{1}{x}+\frac{1}{x+1}\right)=\frac{P(x+1)}{Q(x+1)}-\frac{1}{2}\left(\frac{1}{x+1}+\frac{1}{x+2}\right)$$
Hence it follows by induction that
$$\frac{P(x)}{Q(x)}-\frac{1}{2}\left(\frac{1}{x}+\frac{1}{x+1}\right)=\frac{P(x+n)}{Q(x+n)}-\frac{1}{2}\left(\frac{1}{x+n}+\frac{1}{x+n+1}\right)$$
Fixing $x$ and letting $n \rightarrow \infty$ we see that $\frac{P(x)}{Q(x)}-\frac{1}{2}\left(\frac{1}{x}+\frac{1}{x+1}\right)=c$, where $c$ is a constant. Now it is easy to conclude that $Q(x)=x(x+1) R(x)$ and $P(x)=\left(\frac{1}{2}+x+c x(x+1)\right) R(x)$.