A normed group $(X,+,|\cdot|)$ is a set $X$ equipped with a group operation $+$ and a function $|\cdot|:X\to\Bbb R$ called a norm such that
- $|x|=0\iff x=0$
- $|x-y|\le|x|+|y|$.
From these properties one can also deduce $|x|\ge0$, $|{-x}|=|x|$, and $|x+y|\le|x|+|y|$. (I am using additive notation for the group because the norm's subadditivity property is more consistent with additive notation than multiplicative, but we are not assuming that $+$ is commutative.) These axioms are the necessary and sufficient conditions for $d(x,y)=|x-y|$ to be a metric, and they can also be understood as the intersection of the axioms of an normed $R$-vector space (which adds the axiom $r\in R,x\in X\to|rx|=|r|_R|x|$ where $|\cdot|_R$ is the norm on $R$) and an absolute value over a ring (which adds $|xy|=|x||y|$), but I haven't found any literature directly on normed groups defined like this.
It follows from the definition that $d(x+z,y+z)=d(x,y)$ ($d$ is right translation invariant), but I haven't managed to show $d(z+x,z+y)=d(x,y)$, nor $d(-x,-y)=d(x,y)$ or that the topology induced by $d$ turns $X$ into a topological group, and I think the key is the "identity" $|x+y|=|y+x|$, which I have not managed to prove or disprove.
Claim If $(X,+,|\cdot|)$ is a normed group, then $|x+y|=|y+x|$.
Edit: Turns out this is mentioned in the literature (by which I mean Wikipedia), under the terminology of a length function. Unfortunately I don't see any counterexamples of noncommutative normed groups there which don't satisfy $|x+y|=|y+x|$.
The Wikipedia article tells you how to construct counterexamples: given any presentation of a group $G$ there's an induced length function given by the length of the shortest word in the generators that forms a given $g \in G$ (this is the origin of the term "length function"). There's no reason this should satisfy the condition you want, which is equivalent to length being invariant under conjugation. For example, take $G = F_2$ with the free presentation by two generators $x, y$ and no relations. Then $x$ and $yxy^{-1}$ are conjugate but do not have the same length. The induced metric is also not left translation invariant.