As a toy model for a larger problem, I want to show that if $A,B\geq 0$ and $u(t)$ satisfies
$$|u(t)|\leq A+\left|\int_0^t B\cos(s^2)u(s)ds\right|$$
then $u$ satisfies a bound like
$$|u(t)|\leq AC$$
where $C$ is a (finite) constant depending on integral(s) involving $B\cos(t^2)$. The classical Gronwall lemma is of course not applicable here, since the functions involved are not nonnegative.
For instance, if we try to repeat the proof of the usual Gronwall inequality, then formally we have
$$\frac{d}{dt}\left(A+\left|\int_0^t B\cos(s^2)u(s)ds\right|\right) = B\cos(t^2)u(t)\text{sgn}\left(\int_0^t B\cos(s^2)u(s)ds\right)\leq B|u(t)|$$
After using the bound on $|u(t)|$ we get
$$\frac{d}{dt} \ln\left(A+\left|\int_0^t B\cos(s^2)u(s)ds\right|\right)\leq B$$
Applying the fundamental theorem of calculus and exponentiating yields
$$|u(t)|\leq A+\left|\int_0^t B\cos(s^2)u(s)ds\right| \leq Ae^{Bt}$$
Which is insufficient for my purposes, as I want to make use of the oscillations in the integral of $\cos(s^2)$ to ensure I don't get growth of $|u(t)|$ in time.
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Update: I've found a partial counterexample.
Technically, this counterexample is for a modified problem, but the ideas are similar. Indeed, for any fixed $A,B$ we can find $u$ satisfying
$$|u(t)|\leq A+\left|\int_0^t Be^{is^2}u(s)ds\right|$$
but whose modulus grows exponentially in time (as suggested by the calculation from earlier). Indeed, set $u(t) = Ae^{-it^2}e^{Bt}$. Then we have
$$A+\left|\int_0^t Be^{is^2}u(s)ds\right| = A+(Ae^{Bt} - A) = Ae^{Bt} = |u(t)|$$
It seems harder to come up with an explicit counterexample to the original problem, which suggests the problem may depend on the zeros of the oscillatory kernel.