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
Suppose that the limit of a sum of two functions exists and the limit of one by itself also exists. Then you have a situation where for some real $L$ and $M$ $$\lim_{x \rightarrow a}f(x) + g(x) = L$$ and $$\lim_{x \rightarrow a}f(x)=M.$$ Whenever both limits exist, we can use the difference rule for limits, giving us $$L-M =\lim_{x \rightarrow a}f(x) + g(x) - \lim_{x \rightarrow a}f(x) = \lim_{x \rightarrow a}f(x) + g(x)-f(x)=\lim_{x \rightarrow a} g(x).$$ This proves the limit of the other function must exist as well.