Complete as a semimetric space but not as a topological group I shall begin with some definitions. 
1) Suppose that $X$ is a topological (additive) group and $(x_{s})\subseteq X$ is a net, we said that $(x_{s})$ is Cauchy whenever $U$ is a neighbourhood of $0$, there is some index $r$ such that $x_{s}-x_{t}\in U$ for all $s,t\geq r$. We say that $X$ is complete as a topological group if each Cauchy net converge to some point $x$ in $X$.
2) Suppose that $X$ is a topological group induced by a semimetric $d$, we say that $X$ is complete as a semimetric space with respect to $d$ whenever every sequence $(x_{n})\subseteq X$ is such that $d(x_{m},x_{n})\rightarrow 0$ as $m,n\rightarrow\infty$, we must have some $x\in X$ such that $d(x_{n},x)\rightarrow 0$ as $n\rightarrow\infty$.
Now I am looking for some semimetric $d$ such that $X$ is complete as a semimetric space but not as a topological group. It is known that such $d$ cannot be translation invariant, can anyone suggest any of those? 
 A: I think that there is no such example.
For topological vector spaces this had been a question of Banach and was solved in the fifties by Victor Klee. I think that his arguments also work for topological groups:
Let thus $(X,\tau)$ be a topological group such that $\tau$ is the topology of a complete metric $d$ on $X$. Let $(\hat X,\hat \tau)$ be the completion of
$(X,\tau)$ as a topological group. I am optimistic that $\hat \tau$ is completely metrizable (in the case of topological vector spaces this is true and can be found e.e. in Köthe's book, the proof for groups should be similar: the unit element has a countable neighborhood base which should be used to construct a translation invariant metric. This however needs checking!)
Now, the complete metric space $(X,d)$ is contained as a topological subspace in $(\hat X,\hat \tau)$ and a result of Sierpinski implies that $X$ is a $G_\delta$ set in $\hat X$ which is also dense. Given any $z\in\hat X$ the set
$z-X$ is also dense $G_\delta$ and Baire's theorem implies $X\cap (z-X) \neq \emptyset$, which implies $z\in X+X =X$.
EDIT. The metrizability of topological groups is indeed equivalent to the existence of countable neighborhood bases of the unit element. I found this in Bourbaki's General topology.
