Theorem. For any field $F$, the power series ring $F[[x]]$ is a P.I.D.; in fact, the nontrivial ideals of $F[[x]]$ are of the form $(x^k)$, where $k\in\mathbb{Z}_+$.
Proof. Suppose that $I\subset F[[x]]$ a proper ideal and let $f(x)\in I\setminus\{0\}$ that has minimum order $k$. I have shown that $k$ can not be zero. Then let $k>0$. Therefore \begin{equation} \begin{split} f(x)&=a_kx^k+a_{k+1}x^{k+1}+\cdots+\cdots\\ &= x^k(a_k+a_{k+1}x+a_{k+2}x^{k+2}+\cdots+). \end{split} \end{equation} Question 1. For my book, $x\notin F[[x]]$, so how is the last passage possible?
Question 2. Explicitly who are they $(x^k)$?
Question 3 If you use the convention that $1\cdot x=x$, then is it true or false that even if $x$ is a symbol I can write $x$ how $x=0+1x+0x^2+\cdots+$? And consequently consider $x$ as an element of $F[[x]]$?
(1): By calling those ideals $(x^k)$, we are implicitly saying that there is an element of the ring named $x$, which is exactly what you think it is. Is it the same object as the $x$ that goes into the definition of the ring? Probably not, but we don't really care; we're not going to run into a context where both meanings could fit and would produce different results.
(2): The ideal $(x^k)$ is the set of multiples of $x^k$, exactly as you said. Of course, that means we're multiplying it by any power series, so in terms of coefficients we get $$(x^k) = \left\{a_0x^k + a_1x^{k+1}+\cdots + a_nx^{k+n}+\cdots\ \mid\ a_0,a_1,\cdots\in F\right\}$$ for coefficients up to $k-1$ zero and all others unrestricted.
There's another name that goes with this structure, with one ideal that generates all others as powers of it and everything outside that ideal invertible; this is called a local ring.