Let's consider two independent random variables $$X,Y$$, where $$X\geq0$$. We know both probabilty density functions (pdf) and denote them by $$f_X$$ and $$f_Y$$. We want to compute the pdf of the random variable $$Z:=X\cdot Y$$. In order to do that we take a look at the probability $$P(X\cdot Y\leq z)$$, where $$z\in\mathbb{R}$$.

After a few steps and assuming that $$x>0$$ we get $$\int\limits_{0}^{\infty}\int\limits_{-\infty}^{\frac{z}{x}}f_X(x)f_Y(y)dy~dx.$$ Finally, by taking the derivative we find a pdf of $$Z$$.

The solution in general is clear to me but what is the correct reasoning to say that we can assume $$x>0$$ though it is $$X\geq0$$ in the beginning?

Can I argue like this:

In the context of one dimensional integrals we know that excluding single points doesn't change the integral. I guess that this reasoning can be somehow generalized to two dimensional integrals.

If $$x\leq 0$$ then we know that $$P(X\leq x)=0$$ because of $$X\geq 0$$. Further, we know that $$f_X(x)=(P(X\leq x))'$$ and since $$(P(X\leq X))'=0$$ if $$x\leq 0$$ it follows $$f_X(x)=0$$ if $$x\leq 0$$. So the joint pdf $$f_X(x)f_Y(y)$$ attains $$0$$ if $$x=0$$ and it doesn't matter if we assume $$x>0$$.

Is this correct? Or how would you argue instead if you don't have any further measure theoretic tools at hand?

Your result holds true only if, for example, $$X$$ is a continuous random variable. In general, the solution should be: \begin{align} \mathbb{P}(X\cdot Y\le z) &=\mathbb{P}(X\cdot Y\le z,X>0)+\mathbb{P}(X\cdot Y\le z, X =0)\\ &=\mathbb{P}(X\cdot Y\le z,X>0)+\mathbb{P}(z \ge 0, X =0)\\ &=\color{red}{\mathbb{P}(X\cdot Y\le z,X>0)}+ \mathbf{1}_{\{z \ge 0 \}}\color{blue}{\mathbb{P}(X =0)}\\ \end{align}
The red term is the result in your question : $$\int\limits_{0}^{\infty}\int\limits_{-\infty}^{\frac{z}{x}}f_X(x)f_Y(y)dy~dx$$.
For a $$z \ge 0$$, the blue term can be diffrent to $$0$$ if $$X$$ is a discrete random variable .