First, clear the denominator by multiplying both sides by $x$:
\begin{align*}
z &= \frac{100(x-y)}{x}\\
zx &= 100(x-y)
\end{align*}
Then move all the terms that have an $x$ in it to one side of the equation, all other terms to the other side, and factor out the $x$:
\begin{align*}
zx &= 100x - 100y\\
zx - 100x &= -100y\\
x(z-100) &= -100y
\end{align*}
Now divide through by $z-100$ to solve for $x$; you have to worry about dividing by $0$, but in order for $z-100$ to be $0$, you need $z=100$; the only way for $z$ to be equal to $100$ is if $\frac{x-y}{x}=1$, that is, if $x-y=x$, that is, if $y=0$. Since, presumably, you don't get the things for free, you can assume that $y\neq 0$ so this division is valid. You get:
$$x = \frac{-100y}{z-100}.$$
Now, to get it into nicer form, use the minus sign in the numerator to change the denominator from $z-100$ to $100-z$. Then divide both the numerator and the denominator by $100$ to get it into the form you have:
\begin{align*}
x & = \frac{-100y}{z-100}\\
x &= \frac{100y}{100-z}\\
x &= \frac{\frac{1}{100}\left(100 y\right)}{\frac{1}{100}(100-z)}\\
x &= \frac{y}{1 - \frac{z}{100}}.
\end{align*}
Added: Alternatively, following Myself's very good point, you can go "unsimplify" $\frac{x-y}{x}$ to $1 - \frac{y}{x}$, to go from
$$\frac{z}{100} = \frac{x-y}{x} = 1 - \frac{y}{x}$$
to
$$\frac{y}{x} = 1 - \frac{z}{100}.$$
Taking reciprocals and multiplying through by $y$ gives
\begin{align*}
\frac{x}{y} = \frac{1}{1 - \frac{z}{100}}\\
x = \frac{y}{1-\frac{z}{100}}
\end{align*}
which is probably how the particular expression you had (as opposed to $\frac{100y}{100-z}$) arose in the first place.
When you divide, you are implicitly assuming that the number you are dividing by is not equal to zero. By dividing, you are excluding the possibility that the number in question is zero, and as such you may be eliminating correct answers.
For a very simple example, consider the case of the equation $x^2-x=0$.
There are two answers: $x=0$, and $x=1$. However, if you "divide by the variable", you can end up doing this:
$$\begin{align*}
x^2 - x & = 0\\
x^2 &= x &&\text{(adding }x\text{ to both sides)}\\
\frac{x^2}{x} &= \frac{x}{x} &&\text{(divide by }x\text{, which assumes }x\neq 0)\\
x &= 1.
\end{align*}$$
So you "lost" the solution $x=0$, because when you divided by $x$, you implicitly were saying "and $x\neq 0$". In order to "recover" this solution, you would have to consider "What happens if what I divided by is equal to $0$?"
For a more extreme example, consider something like
$$(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)=0.$$
Since a product is equal to $0$ if and only if one of the factors is equal to $0$, there are six solutions to this equation: $x=1$, $x=2$, $x=3$, $x=4$, $x=5$, and $x=6$. Divide both sides by $x-1$, and you lose the solution $x=1$; divide both sides by $x-2$, you lose $x=2$. Continue this way until you are left with $x-6=0$, and you lost five of the six solutions. And if then you go ahead and divide by $x-6$, you get $1=0$, which has no solutions at all!
Whenever you divide by something, you are asserting that something is not zero; but if setting it equal to $0$ gives a solution to the original equation, you will be excluding that solution from consideration, and so "eliminate" that answer from your final tally.
Best Answer
When you divide by $-2x$ you are implicitly assuming $x\neq0$ which you don't know for sure and thus remove that solution.
Note that when you divide by a non-zero value the equation still holds on each side whereas division by zero will create undefined expressions. For example, $\frac{1}{0}$ doesn't compute to any known value where there are Math SE questions on this topic if you need a reference.
Consider if the equation was $2x=x$ where if I divide by $x$, I then get $1=2$ which is a problem as that isn't true and the reason why that division isn't allowed is because the only solution is $x=0$.