The boundedness of an integral

These integrals are indeed uniformly bounded. As the only effect of changing the signs of $a,b,N$ on the integral is a sign change and/or complex conjugation, we can restrict to $a,b,N\gt0$. Setting $M=Nb$ and $\alpha=a/b^2$, we have $$ \int_{-N}^N\left(e^{i(ax^2+bx)}-1\right)\frac{dx}x=\int_0^M\left(e^{i(\alpha x^2+x)}-e^{i(\alpha x^2-x)}\right)\frac{dx}x. $$ However, the integrand can be written as $2ie^{i\alpha x^2}\frac{\sin x}x$, so is bounded by $2$ in absolute value. Therefore, fixing any constant $K\gt0$, the integral is bounded by $$ \begin{align} 2K+\int_K^{K\vee M}e^{i(\alpha x^2+x)}\frac{dx}x-\int_K^{K\vee M}e^{i(\alpha x^2-x)}\frac{dx}x.&&{\rm(1)} \end{align} $$ I'll prove that this is bounded with the help of a lemma.

Lemma: (van der Corput lemma) If $f\colon[A,B]\to\mathbb{R}$ is convex with $\lvert f^\prime(x)\rvert\ge\lambda\gt0$ and $g\colon[A,B]\to\mathbb{C}$ is differentiable then, $$\left\lvert\int_A^Be^{if(x)}g(x)dx\right\rvert\le\frac2\lambda\left(\lvert g(B)\rvert+\int_A^B\lvert g^\prime(x)\rvert dx\right)$$

As I will only be interested in intervals $[A,B]\subset[K,\infty)$ with $g(x)=1/x$, the bound in the lemma can be written as $$ \begin{align} \left\lvert\int_A^Be^{if(x)}\frac{dx}x\right\rvert\le\frac2{\lambda K}&&{\rm(2)} \end{align} $$ Taking $f(x)=\alpha x^2+x$, this shows that the first integral in (1) is bounded by $2/K$. For the second integral, take $f(x)=\alpha x^2-x$, fix any $0\lt\epsilon\lt1/2$, and first look at value of the integral with integration range restricted to $[0,\epsilon/\alpha]$. In this range, we have $f^\prime(x)\le f^\prime(\epsilon/\alpha)=-(1-2\epsilon)$. So, by inequality (2), this part of the integral is bounded by $2K^{-1}(1-2\epsilon)^{-1}$.

Next, for any fixed $\gamma\gt1/2$, the value of the last integral in (1), restricted to the range $[\epsilon/\alpha,\gamma/\alpha]$ is bounded by $$ \int_{\epsilon/\alpha}^{\gamma/\alpha}\frac{dx}{x}=\log(\gamma/\epsilon). $$ Finally, look at the last integral in (1) restricted to the range $[\gamma/\alpha,\infty)$. As $f^\prime(x)\ge f^\prime(\gamma/\alpha)=(2\gamma-1)$ in this range, inequality (2) shows that this part of the integral is bounded by $2K^{-1}(2\gamma-1)^{-1}$.

Putting these together shows that the set of integrals in the question is bounded above by a constant. The upper bound obtained here is $$ 2K+\frac2K+\frac2{K(1-2\epsilon)}+\frac2{K(2\gamma-1)}+\log(\gamma/\epsilon) $$ for arbitrary positive constants $K,\epsilon,\gamma$ with $\epsilon\lt1/2\lt\gamma$.