Seems like such a simple integral, but I couldn't find the exact expression for the case $a \neq b$.
For $b=a$ we have:
$$\int_0^1 J_0(a r) J_1(a r) dr=\frac{1}{2a} (1- J_0^2(a))$$
Actually, $a,b$ are zeroes of $J_0(x)$, but not necessarily equal. This doesn't seem to help much though.
I couldn't find this integral among the many tables (like this one https://dlmf.nist.gov/10.22).
Numerical values for $a \neq b$ don't seem to lead anywhere either.
This integral comes from a quantum mechanics problem I'm trying to solve. If it doesn't have any nice exact value, I'd rather use a different basis though, despite Bessel functions being otherwise convenient.
Not a full answer.
In the case where $b$ is not too different from $a$, we could use a series expansion $$J_1(b r)=J_1(a r)+\frac{(b-a) (J_1(a r)-a r J_2(a r))}{a}-\frac{(b-a)^2 (r (a r J_1(a r)-J_2(a r)))}{2 a}+$$ $$\frac{r (b-a)^3 \left(a^2 r^2-3\right) J_2(a r)}{6 a^2}+O\left((b-a)^4\right)$$ and integrate termwise to get for the definite integral $$\frac{1-J_0(a){}^2}{2 a}+\frac{a^2 J_0(a){}^2+a^2 J_1(a){}^2+J_0(a){}^2-1}{2 a^2}(b-a)-$$ $$\frac{a^2 J_0(a){}^2+2 a^2 J_1(a){}^2+2 J_0(a){}^2+2}{4 a^3}(b-a)^2+$$ $$\frac{a^4 \left(-J_0(a){}^2\right)-a^4 J_1(a){}^2-2 a^3 J_0(a) J_1(a)+9 a^2 J_0(a){}^2+17 a^2 J_1(a){}^2+18 J_0(a){}^2-18}{36 a^4}$$
Using it with $a=\pi$ and $b=e$, this gives $0.1262693$ while numerical integration leads to $0.1262682$