If $C=cH$, for some subgroup $H$, then you can determine $H$ by taking the set of all values $y^{-1}z$ where $y,z\in C$.
So let's try that. Given a $C$ with the property above, define: $H=\{y^{-1}z: y,z\in C\}$
Step: Prove $H$ is a subgroup - that it is non-empty, and it is closed under multiplication and inverses. (You need to add the condition that $C$ is non-empty to prove this.)
Step: Given any element $c\in C$, all elements can be written as $ch$ for some $h\in H$
So if $C$ has this property and is non-empty then $C$ is a left coset.
If $C$ is a left coset, it is pretty clear it is non-empty, and it's direct to prove that $C$ has this property. Specifically, if $C=cH$, then $x=ch_1$, $y=ch_2$ and $z=ch_3$, for $h_1,h_2,h_3\in H$, and we get $xy^{-1}z = ch_1h_2^{-1}c^{-1}ch_3 = c(h_1h_2^{-1}h_3)\in C$