How separable differential equations can be solved using differentials

Question

My prof defined a separable differential equation as an equation of the form $$N(x) + M(y)\frac{dy}{dx} = 0$$ My understanding is that the goal is to compute a function $\phi: \mathbb{R}^{2} \longrightarrow \mathbb{R}$ such that $\phi(x, y(x)) = 0$ for all $x \in \mathbb{R}$. Of course, if you compute functions $P, Q$ such that $P' = N$ and $Q' = M$ then $$\frac{d}{dx}(P(x) + Q(y(x))) = N(x) + M(y(x))y'(x) = 0$$ so $P + Q \circ y = 0$ and we are done (taking $\phi(u, v) = P(u) + Q(v)$). In lecture, we were told to solve by 'multiplying both sides by $dx$', integrating and rearranging. My understanding of this method is that we consider the $1$-form $$(N + (M \circ y)y')dx$$ which by assumption is identically 0. Here $dx$ is the differential of the identity. Then we expand and use $y'dx = dy$ (which I understand is fully rigorous) to get \begin{align*} 0 = Ndx + (M\circ y)dy = dP + d(Q \circ y) = d(P + Q \circ y) \end{align*} to arrive at the same answer as before. My question is; is this really what is going on when we multiply by dx? If there is a different way to do this rigorously, please write it as an answer. The notation that my prof used makes me think that there's something deeper going on. He writes \begin{align*} 0 &= N(x)dx + M(y)dy\\ -\int N(x)dx &= \int M(y)dy\\ -P(x) &= Q(y)\\ \end{align*} so $0 = P(x) + Q(y)$ Is this abuse of notation (or flat out not rigorous) or am I just failing to see how this is all correct? Is the expression $\int M(y)dy$ even defined, or is it implicitly a shorthand for $\int (M \circ y)dy$? In general, I feel like there are a bunch of subtleties with regards to the notation (subtleties that make the notation much more suggestive). I know its vague but if someone can explain the notation that is commonly used and why its used/why its elegant, that would be greatly appreciated.