Let $G$ be a finite group. Define $d:G\times G\longrightarrow\mathbb{N}$ by $d(x,y)=o(xy^{-1})-1, \forall\, x,y\in G$. Then $d$ is a metric on $G$ if and only if
$$(\*)\hspace{5mm}o(ab)<o(a)+o(b), \forall\, a,b\in G.$$It is easy to see that an abelian group satisfies $(\*)$ if and only if it is a $p$-group. Moreover, if an arbitrary group $G$ satisfies $(\*)$, then all its elements must be of prime power order, i.e. $G$ must be a $CP$-group (a classification of these groups can be found here: <http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.45.3738&rep=rep1&type=pdf>). I have some difficulties to check whether a $CP$-group satisfies $(\*)$. Is there another way to proceed? Also, is it interesting to study such a metric on finite groups?


Additional question: Is it true that all finite groups satisfying $(\*)$ are solvable? By the answer below it follows that this class contains no nonabelian simple CP-group (in fact, no nonabelian simple group). Is it possible that a group in the other class (ii) of nonsolvable CP-groups to satisfy $(\*)$? 



On a metric associated to certain finite groups
The group ${\rm PSL}(2,7)$ seems to be a $CP$ group according to your definition, but it contains an element $a$ of order $2$ and an element $b$ of order $3$ whose product $ab$ has order $7,$ so you do not have a metric in this group.
 In fact, as I write, I realise that in a group $G$ where $d$ defines a metric, the product of any two elements of order $2$ can have order at most $3.$ This seems to exclude $G = {\rm PSL}(2,q)$ as a possibility whenever $q >3.$ Also, the Suzuki groups (the only non-Abelian simple groups of order prime to $3$) are excluded, as they always contain a dihedral subgroup of order $10.$


Later edit: note that the structure of a Sylow $2$-subgroup $S$ of a group $G$ such that $d$ defines a metric is extremely restricted. The elements of order at most $4$ in $S$ form a subgroup, and all involutions of $S$ commute with each other. Thus $S$ has no dihedral subgroup of order $8$ and no generalized quaternion subgroup of order $16$ or more. Even later edit: note also that by an earlier remark, if $t$ and $u$ are involutions of $G,$ then either $t$ and $u$ are conjugate (via an element of order $3$), or else $t$ commutes with every $G$-conjugate of $u.$


