For every $\varepsilon>0$ find a piecewise continuous function $q:[0,1]\rightarrow \mathbb{R}$ such that $\int\_0^1 q(x)dx=1$ and
$$\int\_{0}^1 \int\_{0}^{s} \left|\frac{q(s)q(t/s)}{s}- \frac{q(t)q((s-t)/(1-t))}{1-t}\right|dt ds<\varepsilon$$
Somehow the question does not look hard, but still I don't see if it is true or not. 
For the function $1/(x-x^2)$ one has $\frac{q(s)q(t/s)}{s}- \frac{q(t)q((s-t)/(1-t))}{1-t}=0$, but the truncation of this function to $[\varepsilon,1-\varepsilon]$ and normalization do not work.



Approximation of an integral over the unit ball of L_1
There does not exist such a function $q$ if $\varepsilon<1/2$.


Indeed, if $q$ is a positive measurable function on $[0,1]$ of integral $1$, pick $a \in [0,1]$ such that $\int\_0^a q(x) dx = 1/2$.


Then 
$$\int\_{0}^a \int\_{0}^{s} \frac{q(s)q(t/s)}{s} dt ds = 1/2,$$ 
whereas
$$\int\_{0}^a \int\_{0}^{s} \frac{q(t)q((s-t)/(1-t))}{1-t}dt ds$$ is equal to 
$$ \int\_{0}^a \int\_{0}^{\frac{a-t}{1-t}} q(t) q(u) du dt,$$
But since $(a-t)/(1-t) \leq a$ for all $t\leq a$, this quantity is less than $1/4$.


This implies that 
$$\int\_{0}^1 \int\_{0}^{s} \left|\frac{q(s)q(t/s)}{s}- \frac{q(t)q((s-t)/(1-t))}{1-t}\right|dt ds \geq 1/2.$$


