Why { $z-x-y=0$ , $z-2x=0$ , $2x+y-3z=0$ } cannot be solved this way?

Your friend used just two equations, which (since they were linearly independent) is enough to limit the solution space to one dimension. The correct conclusion is not that $\langle x,y,z\rangle = \langle 1,-\frac12,\frac12\rangle$ is the solution, but that the solution has the form $$\langle x,y,z\rangle = \left\langle t,-\frac12t,\frac12t\right\rangle$$ for some real number $t.$

This is true. The correct solution does have that form.

Now compare this with your solution and see if you can tell what value(s) of $t$ give a solution to all three equations.

You wrote, correctly, that the second equation was not used at all. And the solutions of the system which consists of the other two equations are the triplets $(x,y,z)$ such that $x=-2y$ and that $z=-y$ indeed. Finally, there is not reason at all to pick $x=1$.

Let me present an extreme version of your friend's argument:

By the first equation, we have

$$ z = x + y $$

Then, simply choose $ x = y = 1 $ and conclude $ z = 2 $.

Thus, $(1,1,2)$ is a solution to the system!

Or to a ridiculous degree:

By ignoring all equations, we can choose $x = y = z = 1$ and so $(1,1,1)$ is a solution!

I hope this illuminates the issue with your friend's solution. Each equation potentially restricts the possible solutions to the system, and so each equation not used potentially un-restricts those solutions to ones which are not actually valid.

One more example to demonstrate this fact:

Consider the system of two equations for one variable, $$ \begin{align*} x &= 1 \\ x &= 2 \end{align*} $$ By limiting our scope to only one equation, we may conclude either that $1$ is a solution or that $2$ is a solution, but in reality there are (obviously) no solutions, and so both are wrong.

We can begin by substituting $(2)$ into $(1)$ or $(3)$. The problem is, using $(2)$, we find a contradiction between $(4)$ and $(6)$

\begin{align} z-x-y=0 \implies -x-y+z&=0 \tag{1}\\ z-2x=0 \implies z&=2x \tag{2}\\ 2x+y-3z=0\implies 2(x)+y-3z&=0 \tag{3} \end{align} Substitute $(2)$ into $(1)$ then $(4)$ into $(3)$

\begin{align} -x-y+(2x)=0\implies x&=y \tag{4}\\ 2(x)+y-3z=0\implies 3y-3z&= 0\tag{5}\\ \implies y=z\implies y&=2x \tag{6} \end{align}

or we can combine $(1)$ and $(3)$ as you did.

Sometimes rearranged things are earier to "see" \begin{align} z-x-y=0 \implies -x-y+z=0 \tag{1}\\ z-2x=0 \implies -2x+0y+z=0 \tag{2}\\ 2x+y-3z=0\implies 2(x)+y-3z=0 \tag{3} \end{align} Add $(1)$ to $(3)$, substitute $(4)$ into $(3)$ then $(5)$ into $(6)$, etc.

\begin{align} x-2z=0\implies x&=(2z) \tag{4} \\ \implies z&=\frac{x}{2} \tag{5}\\ 2(2z)+y-3z=0\implies y&=(-z) \tag{6}\\ \implies y&=\frac{-x}{2} \tag{7} \end{align}

Setting $x=1\quad \bigg(x,\dfrac{-x}{2},\dfrac{x}{2}\bigg) =\bigg(1,\dfrac{-1}{2},\dfrac{1}{2}\bigg) $