For any $1 \leq p<q$ we have:
$$\frac{p}{q}=\frac{1}{q^2-p q+1}+\frac{(q-p)(pq-1)}{q(q^2-p q+1)}$$
Let's consider an iteration:
$$p_{n+1}=(q_n-p_n)(p_nq_n-1)$$
$$q_{n+1}=q_n(q_n^2-p_n q_n+1)$$
Then we have: $$\frac{p_0}{q_0}=\sum_{n=0}^\infty \frac{1}{q_n^2-p_n q_n+1}+\lim_{n \to \infty} \frac{p_n}{q_n}$$
The above is trivially true, provided both limits exist separately. But it turns out numerically that both the sum and the limit are finite and (seemingly) irrational. Let's denote:
$$A(p_0,q_0)=\sum_{n=0}^\infty \frac{1}{q_n^2-p_n q_n+1}$$
$$B(p_0,q_0)=\lim_{n \to \infty} \frac{p_n}{q_n}$$
$$A(p_0,q_0)+B(p_0,q_0)=\frac{p_0}{q_0}$$
We have (I will only write $A$, since $B$ can be obtained by subtraction):
$$A(1,2)=0.365624790175942982737859474249681505 \dots$$
$$A(1,3)=0.145650460727123812176794888179825955 \dots$$
$$A(2,3)=0.261766321023330525617942741920174815 \dots$$
$$A(3,4)=0.205525029400010131449324836277780238 \dots$$
Moreover, since we define 'numerators' and 'denominators' separately, we obtain different results for $A(mp_0,mq_0)$:
$$A(2,4)=0.1123721471627326977547524155555190359 \dots$$
$$A(3,6)=0.0527707965471706424044987365216530762 \dots$$
$$A(4,8)=0.0303300858480840619918350602972694890 \dots$$
$$A(4/3,8/3)=0.228842771329071793825717533784828873 \dots$$
I haven't been able to find a closed form for any of $A,B$ I tried.
However, it seems likely they are irrational. If it's true and there is no known closed form, then we can produce an infinite number of pairs of irrationals $A,B$ which sum to a particular rational number nontrivially.
My questions are:
Can we prove that $A,B$ are irrational for rational $p_0,q_0$?
Is there any closed form for $A,B$ in terms of $p_0,q_0$?
Are the values of $A,B$ unique for every distinct pair of $p_0,q_0$?
A sketch of a proof of existence for $B$ (note that we assume $p_0,q_0 \in \mathbb{N}$ and $1 \leq p_0 \leq q_0-1$):
$$p_{n+1}=(q_n-p_n)(p_nq_n-1) \geq p_n^2+p_n-1$$
$$q_{n+1}=q_n+q_n^2(q_n-p_n) \geq q_n^2+q_n$$
We can see that $p_n,q_n$ are non-decreasing, and for $n>1$ they are increasing (because $q_n$ is strictly increasing and it 'helps' $p_n$ after the first step).
For $n \to \infty$ we have $p_n \to \infty$ and $q_n \to \infty$, thus:
$$\frac{p_{n+1}}{q_{n+1}}=\frac{(q_n-p_n)(p_nq_n-1)}{q_n(q_n^2-p_n q_n+1)} \approx \frac{p_n}{q_n}$$
It is apparent the limit exists.
The limit for $A$ exists because the sequence $q_n^2-p_n q_n+1$ grows much faster than $n^2$ and the sum obviously converges.
Update
A little something on a closed form. The system of recurrence relations can be rewritten as a single recurrence relation, using:
$$p_n=q_n+\frac{1}{q_n}-\frac{q_{n+1}}{q_n^2}$$
Then we have a second order recurrence relation:
$$q_{n+2}=q_{n+1}(q_{n+1}q_n+1)+\frac{q_{n+1}^3}{q_n^2} \left(\frac{q_{n+1}}{q_n}-1 \right)$$ $$q_0=q_0, \qquad q_1=q_0(q_0^2-q_0 p_0+1)$$
Or a more symmetric form:
$$\frac{q_{n+2}}{q_{n+1}}=q_{n+1}q_n+1+\frac{q_{n+1}^2}{q_n^2} \left(\frac{q_{n+1}}{q_n}-1 \right)$$
If we find a closed form for it (which I'm not sure exists) we can take the limit and find the closed form for $B$.
We also have a more simple looking relation (but it still requires us to know $q_n$):
$$\frac{p_n}{q_n}=1+\frac{1}{q_{n-1}^2}-\frac{q_{n-1}}{q_n}-\frac{q_n}{q_{n-1}^3}$$
And in fact, we can also write $A$ in terms of $q_n$:
$$A=\sum_{n=0}^\infty \frac{q_n}{q_{n+1}}$$
Update 2
Getting rid of some unnecessary parts, we can reformulate the problem:
Set some $q_1>q_0>0$. Then we can define a second order recurrence:
$$q_{n+2}=q_{n+1}(q_{n+1}q_n+1)+\frac{q_{n+1}^3}{q_n^2} \left(\frac{q_{n+1}}{q_n}-1 \right)$$
With the following property: $$L(q_0,q_1)-S(q_0,q_1)=\lim_{n \to \infty} \frac{q_{n+1}}{q_n^3}- \sum_{n=0}^\infty \frac{q_n}{q_{n+1}}=\frac{q_1-q_0}{q_0^3}$$
Can we find a closed form for the recurrence? Or separately for the limit $L$ or the sum $S$ above?
Note that for the limit $L$ to be finite we need to have as $ n \to \infty$:
$$q_n \asymp C \cdot a^{3^n}$$
For example we have:
$$S(1,2)=0.645953147800624278311945190231458547= \\ = \frac{1}{2}+\frac{1}{7}+\frac{1}{323}+\frac{1}{33657247}+\frac{1}{38127274806076464952763}+\dots$$
No closed form for this number either, however look at the denominator sequence - all the numbers end with $3$ or $7$. This pattern continues as far as I can see.