This fraction is at least $\frac{1}{k(G)}$ where $k(G)$ is the number of conjugacy classes of $G,$ because there are $|G|^{2}$ expressions of the form $[a,b]$ and every element of $G$ has at most $k(G)|G|$ expressions of the form $[a,b]$. Hence the number of distinct commutators in $G$ is at least $\frac{|G|^{2}}{k(G)|G|} = \frac{|G|}{k(G)}.$

On the other hand a non-Abelian extraspecial $2$-group $G$ of order $2^{2n+1}$ has only $2$ commutators, and has $k(G) = 2^{2n} + 1,$ so $\frac{|G|}{k(G)} \to 2$ as $n \to \infty,$ while we always have $\frac{|G|}{k(G)} < 2.$

The proportion of elements of a finite group $G$ which are commutators is at least $\frac{1}{k(G)}$ (and the inequality is strict for non-Abelian groups) where $k(G)$ is the number of conjugacy classes of $G,$ because there are $|G|^{2}$ expressions of the form $[a,b]$ and every element of $G$ has at most $k(G)|G|$ expressions of the form $[a,b]$ (and only at the identity can that bound be achieved). Hence the number of distinct commutators in $G$ is at least $\frac{|G|^{2}}{k(G)|G|} = \frac{|G|}{k(G)}.$

On the other hand (apart from the obvious example of Abelian groups where the boound is clearly sharp), a non-Abelian extraspecial $2$-group $G$ of order $2^{2n+1}$ has only $2$ commutators, and has $k(G) = 2^{2n} + 1,$ so $\frac{|G|}{k(G)} \to 2$ as $n \to \infty,$ while we always have $\frac{|G|}{k(G)} < 2.$ Hence the proportion of commutators can get arbitrarily close to $\frac{1}{k(G)}$ for non-Abelian groups