We know that du Bois-Reymond lemma (also known as the fundamental lemma of variational calculus) states that, given $\int_{a}^{b} f(t) \dot{h}(t) dt = \int_{a}^{b} g(t) h(t) dt$ for all $h$ compactly supported and smooth, then $\dot{f}(t) = g(t).$
Are there any analogs for that if, for example, $h$ is integrated: $$\int_{a}^{b} f(t) \dot{h}(t) dt = \int_{a}^{b} \left( g(t) h(t) + \int_{a}^{t} e(\theta)h(\theta) d \theta \right) dt$$
Answering my own question here in case someone may find it helpful.
So you just have to use Fubini's theorem to swap to integrals like this: $$\int\limits_a^b \int\limits_a^t e(\theta) h(\theta) \, d\theta \, dt = \int\limits_a^b \int\limits_\theta^b e(\theta) h(\theta) \,dt \, d\theta = \int\limits_a^b (b-\theta) e(\theta) h(\theta) \, d\theta = \int\limits_a^b (b-t) e(t) h(t) dt.$$
Thus you have only left the integrals by $dt$ on the interval $[a,b]$. Proceed to apply du Bois-Reymond lemma.