Where is the mistake in this logarithmic equation? The problem is:

Given this equation find the possible values of $x$.
  $$\log_{2}\left ( \frac{x+2}{x-1} \right )+\log_{2}\left [ (x+2)(x-1) \right ]=2$$

First I defined the domain of the function.


*

*The denominator $x-1$ must not be zero: $x-1\neq0$.

*The fraction $\frac{x+2}{x-1}$ must be bigger than zero: $\frac{x+2}{x-1}>0$.

*The product $(x+2)(x-1)$ must be bigger than zero: $(x+2)(x-1)>0$.
In the end I concluded that the domain should be $\left ]-\infty,-2\right [\cup \left ]1,+\infty\right[$.
Then I sarted to solve:
$$\begin{align*}
\log_{2}\left ( \frac{x+2}{x-1} \right )&+\log_{2}\left [ (x+2)(x-1) \right ]=2 
\\
&\iff
\log_{2}(x+2)-\log_{2}(x-1)+\log_{2}(x+2)+\log_{2}(x-1)=2\\
&\iff
2 \cdot \log_{2}(x+2)=2\\
&\iff \log_{2}(x+2)=1 \\
&\iff x+2=2^1\\
&\iff x=2-2\\
&\iff x=0.
\end{align*}$$
This result is not in the domain!
Where I miss?

One problem is that you can have both the product and the fraction be valid arguments for $\log_2$, but the individual factors/denominator/numerator not be valid arguments. Specifically, if $x\lt -2$, then neither $x-1$ nor $x+2$ are valid arguments, so you cannot take the first step in general.
Instead, why not work those identities in the other direction? 
$$\begin{align*}
\log_2\left(\frac{x+2}{x-1}\right) + \log_2\Bigl( (x+2)(x-1)\Bigr) 
&= \log_2\left(\left( \frac{x+2}{x-1}\right)(x+2)(x-1)\right)\\
&= \log_2\left(\bigl(x+2\bigr)^2\right).
\end{align*}$$
Now, if $x+2\gt0$ (and hence, if $x\gt 1$, then this is equal to $2\log_2(x+2)$, from which you conclude that $\log_2(x+2) = 1$, hence $x+2=2$, hence $x=0$, but that is not in the domain. 
What does that mean? It means that you cannot find a solution with $x+2\gt 0$.
What if $x+2\lt 0$, though? Then simply go from $\log_2((x+2)^2) = 2$ to $(x+2)^2 = 2^2 = 4$, and hence $-(x+2) = 2$, or $x+2=-2$, hence $x=-4$.  This is in the domain, and satisfies $x+2\lt 0$.
Indeed, note that if $x=-4$, then $\frac{x+2}{x-1} = \frac{-2}{-5} = \frac{2}{5}$, and $(x+2)(x-1) = (-2)(-5) = 10$, so you have that the left hand side is
$$\log_2(2/5) + \log_2(10) = \log_2(20/5) = \log_2(4) = 2,$$
so $x=-4$ is a solution.
