Neutral element in monoid. Let  $(G, *) $ be a monoid . Let $g \in G$ and $ g = g * g $. Can I assume, that $g$ must be neutral element? Why?
 A: In fact there are monoids all of whose elements have this property (which is called being idempotent). For example, you can take the underlying set of any poset with finite joins or finite meets (semilattices). 
As an example of that example, the totally ordered set $\{ 1, 2, \dots n \}$ has joins and meets given by taking max and taking min respectively. Both of these define a monoid structure where every element is idempotent. 
A: No. Take $G=\mathbb Z$ and $g=0$. 
Far less drastic examples can be found in matrices. 
A: Minimal counterexample. The monoid $\{0, 1\}$ with the usual multiplication of integers. Then $0 * 0 = 0$ and $1 * 1$ but $0 \not= 1$. 
A: No, and here's a more complex example:  let $\Bbb N$ be the natural numbers, i.e. $\Bbb N = \{ 0, 1, 2, 3, \ldots \}$; note I am allowing $0 \in \Bbb N$.  For any $n \in \Bbb N$, $n \ge 2$, let $M_n(\Bbb N)$ be the set of $n \times n$ 
matrices with entries taken from $\Bbb N$.  It is clear that $M_n(\Bbb N)$, with the usual operation of matrix multiplication, forms a (non-commutative) monoid with identity $I$.  However, any matrix $G$ formed by replacing at least one diagonal entry of $I$ with $0$ also satisfies $G^2 = GG = G$, but it is easy to construct examples showing no such $G$ satisfies $GA = AG = A$ for all $A \in M_n(\Bbb N)$; e.g. take $A = I - G \ne 0$ for some such $G$.  Then $AG = GA = G(I - G) = G - G^2 = 0 \ne A$.  Such $G$, though idempotent, are not the identity element of $M_n(\Bbb N)$.
A: No, you cannot.  $\mathbb{N}^{\geq 0}$ is a monoid under multiplication, but $g=0$ is not the identity element even though it satisfies your condition.
A: As pointed out already by five answers before mine, no. It is easy to construct the Cayley table of a finite counterexample. Let $G=\{1,2,3\}$, then
$$\begin{array}{c|ccc}
\ast & 1 & 2 & 3\\\hline
1 & \color{red}{1} & 2 & 3\\
2 & 2 & \color{red}{2} & 3\\
3 & 3 & 3 & \color{red}{3}\\
\end{array}$$
defines a commutative (idempotent) monoid $(G,\ast)$ where $2\ast 2=2$ and $3\ast 3=3$ but the neutral element is $1$.
