Is$ \prod_{n=1}^{\infty} (x-n\pi)$ a factor of $\sin x$? As it holds for a polynomial it should be true for a power series too. Why not?
The product is not convergent for $x$ except for positive integer multiples of $\pi$, where it is $0$. So given that it isn't even well defined for most of $\mathbb{R}$, we shouldn't expect it to be a factor of $\sin x$.

A power series is not necessarily a polynomial, nor is an infinite product of factors. And just because something holds in the finite case does not mean it holds in the infinite case. For example, all polynomials have derivatives that are also polynomials, so should we expect this to be true in the limit with power series of $e^x$ or $\sin x$? No, so we shouldn't expect an infinite product of $(x-r_i)$ for roots $r_i$ of a function to equal that function. In fact, why should we expect your product to equal $\sin x$ and not any of the other functions with the same roots like a triangle wave or a square wave?

Euler managed to make the product work by manipulating it into a convergent product but there is more going on behind the scenes than it seems. There is more detail on that product in (Question 674769).
No, because this product does not converge. e.g. for $x = 0$ you get $\pi \cdot 2\pi \cdot 3\pi \rightarrow \infty$.  You also forgot about the negative roots. But you can use this general idea of factoring $\sin(x)$ into linear factors corresponding to the roots.  If you write the linear factors as $1 - x/(\pi n)$, with $x=\pi n$ as the corresponding roots, and group the roots in pairs $\pm \pi n$ to get $$(1 - \frac{x}{\pi n})(1 + \frac{x}{\pi n}) = 1 - \frac{x^2}{\pi^2n^2},$$ and divide away the factor $x$ corresponding to the root $x = 0$, you might guess that $$\frac{\sin(x)}{x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{\pi^2n^2}\right)$$ and this turns out to be true! That is, the right-hand product converges to $\sin(x)/x$ for all $x$.  This was used in Euler's famous first solution to the Basel problem, expressing $\sin (x) /x$ as a Taylor series and extracting the $x^2$ term on both sides to get $$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$$ Not simple to prove this factorization works, though, without more analysis.  In particular this is justified by the Weierstrass factorization theorem, proved some hundred years later.  But how did Euler know that this would work?  Well, he was Euler.  A genius, and also a bit fast and loose with his logic.