I have seen that in option pricing problems (Black-Scholes (BSE)), various asset price dynamics $S_t$ are assumed to derive models. For the Generalized Black-Scholes equation (GBSE), we can see that the change in stock price $dS$ of the underlying satisfies the stochastic differential equation $$dS = (\mu- D)S dt + \sigma S dW,$$ where $μ$ is the drift rate, $D$ is the dividend yield, $σ$ is the market volatility, and $dW$ is the increment of a standard Wiener process. How can we treat the above equation as a 'differential equation' since the increment $dW$ is taken from a mere continuous function (random variable)? How the so-called 'differential equation' is sensible throughout the genesis of a partial differential equation(GBSE)?
Further, In some models like Jump-diffusion BSE, Jump discontinuity is even assumed in the process. How do we justify the analysis (real) of such models?
I'm going to explain this in terms a sequence of steps of increasing generality.
N.b.: There are some technical flaws in my presentation below. It's been a while since I learned this, and I clearly misremember some details. See Paresseux Nguyen's comment below, which cites the standard texts for the argument. I haven't yet digested exactly how to fix my mistakes; once I do, I'll try to correct the presentation.
1. Replace the differential equation with an integral equation
You seem to implicitly assume that $$dS=(\mu-D)S\,dt+\sigma S\,dW$$ is shorthand for "the equation $$\frac{dS(t)}{dt}=(\mu-D)S(t)+\sigma S(t)\frac{dW(t)}{dt}$$ holds for any $t$." Instead, interpret it as shorthand for "the equation $$\int_{t=0}^{t=T}{dS(t)}=\int_0^T{(\mu-D)S(t)\,dt}+\int_{t=0}^{t=T}{\sigma S(t)\,dW(t)}$$ holds for any $T$."
If everything is smooth, then both representations are equivalent by the fundamental theorem of calculus. Of course, not everything is smooth; that's the thrust of your question! It turns out the definition of an integral tolerates rough functions much better than that of a derivative.
2. The Riemann-Stieltjes-Young integral
Given functions $f$ and $g$, define (when it exists) $$\int_{t=0}^T{f(t)\,dg(t)}=\lim_{\lambda\vdash[0,T]}{\sum_{[y,z]\in\lambda}{f(x_{[y,z]})(g(z)-g(y))}}$$ where the limit is over further refinement of the partition $\lambda$ and should be invariant over all possible choices of $x_{[y,z]}\in[y,z]$. One can then show:
Of course, functions of bounded variation are differentiable a.e., so Brownian motion $W$ cannot be of bounded variation.
OTOH, $W$ is $\frac{1}{2-\epsilon}$-Hölder. As it turns out, that's enough.
3. Rough paths theory
Define the $p$-variation of a function $f$ on $[0,T]$ to be $$V_p(f,[0,T])=\lim_{\lambda\vdash[0,T]}{\sqrt[p]{\sum_{[x,y]\in\lambda}{|f(y)-f(x)|^p|}}}$$ with the obvious modifications if $p=\infty$. A function is bounded if it has finite $\infty$-variation and of bounded variation if it has finite $1$-variation. Brownian motion has finite $2$-variation.
From Hölder's inequality, if $f$ has finite $p$-variation, $g$ has finite $q$-variation, and $\frac{1}{p}+\frac{1}{q}=1$, then there exists a Riemann-Stieltjes-Young integral $$\int_{t=0}^T{f(t)\,dg(t)}$$
As mentioned, Brownian motion has finite $2$-variation, as does the solution to any Brownian SDE. Thus the integrals in $$\int_{t=0}^{t=T}{dS(t)}=\int_0^T{(\mu-D)S(t)\,dt}+\int_{t=0}^{t=T}{\sigma S(t)\,dW(t)}$$ all exist in the Riemann-Stieltjes-Young sense.