Let $G$ be a group and $X$ be a G-set.
If $a \cdot x= b\cdot x$ for some $x \in X$, then $a=b$ (in G).
Is this true? I don't fully understand the notion of an 'action'. One part of me thinks this isn't true as we don't know if $x$ has an inverse. But another part of me thinks this could be true if the action is multiplication on a certain set (such as $\mathbb{Z}$).
I can certainly see that this could be false if the action was multiplication on $\mathbb{Z_4}$.
Since $x$ is not an element of the group, it doesn't make sense to say it has an inverse.
Consider the symmetry group of a triangle $\triangle ABC$. A rotation may move $A$ to $B$, but so does the reflection across the line through $C$, and yet the rotation is not the same as the reflection.
Or consider the symmetry group of a sphere. Any two rotations around the vertical axis will fix the north pole (i.e. $R\cdot N=N$ where $R$ is the rotation and $N$ is the north pole) but these rotations can be by different angles and so can be different group elements.
The condition that $ax=bx~\Rightarrow a=b$ for all $a,b\in G$ and $x\in X$ is that $G$ acts semiregularly. One may logically equate $ax=bx$ with $(b^{-1}a)x=x$, so the action being semiregular is equivalent to no (nontrivial) group element having a fixed point.