While in some contexts,
$$
\frac{d\textbf{p}}{dt}dt = d\textbf{p}
$$
is correct and is mathematically rigorous, there is a straightforward way to derive impulse as the change in momentum. Consider a function $f(x)$ where $f:\mathbb{R}^m \to \mathbb{R}^n$ and $f \in C^1$. Suppose its derivative is $g = f'(x)$. We can consider integrating $g(x)$ over some interval $I = [x_i, x_f] \subset \mathbb{R}^m$ such that
$$
\int_I g(t)dt = \int_{x_1}^{x_2}g(t)dt = f(x_2) - f(x_1),
$$
by the fundamental theorem of calculus. We can apply this result to momentum to get your result. Consider that the momentum $\textbf{p}(t)$ of a particle is simply a function $\textbf{p}: \mathbb{R} \to \mathbb{R}^3$. The impulse is defined as the net external force $\textbf{F}_{ext}(t)$ that a particle experiences over some time interval $I = [t_1, t_2]$, or
$$\textbf{J} \equiv \int_I \textbf{F}(t)dt,$$
where the "ext" has been dropped for brevity. (It is still understood that it represents the net external force.) By definition, $\textbf{F}(t) \equiv \textbf{p}'(t),$ so we can apply the fundamental theorem of calculus to obtain the desired result:
$$\int_{t_1}^{t_2} \textbf{F}(t)dt = \textbf{p}(t_2) - \textbf{p}(t_1) = \Delta \textbf{p}.$$