The first thing I try to do in situations like this is to sketch a graph of the function in the neighborhood where the limit is evaluated. In your case, it is not hard to see that when $x$ is sufficiently small and positive, $\lfloor \sin x \rfloor = 0$, for example. So then $$f(x) = \frac{x}{x- \lfloor \sin x \rfloor}$$ in this region of $x$ will simply be equivalent to $x/x = 1$.
But for $x < 0$ but sufficiently close to $0$ (e.g., $x = -0.1$), $\lfloor \sin x \rfloor$ is no longer zero, but $-1$, because you are "rounding down." Therefore, $f(x) = x/(x+1)$ in this region, and you can sketch the graph accordingly.
Now you can see that there is a jump discontinuity at $x = 0$. Moreover, you can see that the discontinuity is of size $1$, which is formally represented by the fact that $$\left| \lim_{x \to 0^+} f(x) - \lim_{x \to 0^-} f(x) \right| = 1.$$ That suggests a choice of $\epsilon < 1$ will work to prove the two-sided limit does not exist, since if, say, $\epsilon = 1/2$, you will not be able to find any $\delta$ such that the difference $|f(x) - L| < 1/2$ in the neighborhood of $x = 0$. To formalize this reasoning is an exercise I leave to you.
Question 1: Are there some conditions where we can somehow say that the limit of the product doesn't exist because limit of 1 of the factors doesn't exist?
Yes, indeed you can. Frequently, if the limit of exactly one of the two terms exists, then their product (or sum) will also fail to exist, but there are caveats. Formally, if $\lim_{x\to a} f(x)$ exists and is non-zero, and $\lim_{x \to a} g(x)$ does not exist, then $\lim_{x \to a} f(x)g(x)$ does not exist too.
Proving this is a straightforward proof by contradiction. We can write
$$g(x) = \frac{f(x)g(x)}{f(x)}.$$
If $\lim_{x \to a} f(x)g(x)$ exists, then by the algebra of limits, given once again the non-zero limit of the denominator, then the limit of $g(x)$ would exist and we would have
$$\lim_{x \to a} g(x) = \frac{\lim_{x \to a} f(x)g(x)}{\lim_{x \to a} f(x)},$$
which contradicts the limit of $g$ not existing. So, our assumption that $\lim_{x \to a} f(x) g(x)$ exists is wrong.
A similar result holds for $f(x) + g(x)$, except we need not assume $\lim_{x \to a} f(x) \neq 0$.
In your example, $\frac{x}{1} \cdot \frac{1}{x}$, the problem is, of course, that your $f(x) = \frac{x}{1}$ tends to the one and only forbidden limit: $0$. This provides a suitable counterexample to show that the $\lim_{x \to a} f(x) \neq 0$ condition is necessary.
I initially tried arguing$$\lim_{(x,y) \to (0,0)}e^{\frac{y}{x^2+y^2}}\cos\left(\frac{x}{x^2+y^2}\right)$$doesn't exist because $\lim_{(x,y) \to (0,0)} e^{\frac{y}{x^2+y^2}}$ doesn't exist...
Question 2: But this is incorrect reasoning...right?
This is indeed incorrect, as the limit of neither term exists. Instead, try considering the limit along the two axes.
Best Answer
Depends on the context in which you're working.
If it's with the reals, for example, then such limits simply fail to exist. However, we do sometimes work with the extended reals $[-\infty,+\infty],$ and clearly in that case we can say something like $\lim_{x\to+\infty}x=+\infty.$