Are these special (one production) Context-Free Grammars always unambiguous?
Here is a simple counter example:
$S \rightarrow aSbSaSbS \space |\space  \epsilon$ 

and string $w: abababab.$ 
In one case we use last $S$ and in other case we use second $S$. All other $S$ goes to $\epsilon$.



Why I was able to get this grammar?

Let's rewrite above grammar with numbers assigned to each $S$'s.

$S \rightarrow aS_1bS_2aS_3bS_4 | \epsilon$

Now We can virtually eliminate $S_1$ and $S_3$. (For this whenever we have to derive something from $S_1$ or $S_3$ we will only derive $\epsilon$ from both of this type of $S$'s.)

So we get $S \rightarrow abS_2abS_4|\epsilon$.  (At this time we already have got rid of rule 3.)

Which we were looking for.