embedding of abelian topological group.

Let $A=\mathbb{Z}$, with the topology given by embedding it in the product $\prod_n \mathbb{Z}/(2\cdot 3^n\mathbb{Z})$ by the quotient map on each coordinate. (So, we have the $3$-adic topology restricted to $2\mathbb{Z}$, but $2\mathbb{Z}$ is open in $\mathbb{Z}$.) Then multiplication by $2$ is injective on $A$, but it is not an embedding, because $2\mathbb{Z}$ is open in $A$ but its image $4\mathbb{Z}$ is not.

There's a locally compact example: fix a prime $p$, and consider $\mathbf{Q}_p^\mathbf{N}$, endowed with the unique topology for which $\mathbf{Z}_p^\mathbf{N}$, with the (compact) product topology, is an open subgroup. (This topology is induced by the diagonal inclusion into $\mathbf{Q}_p^\mathbf{N}\times (\mathbf{Q}_p^\mathbf{N}/\mathbf{Z}_p^\mathbf{N})_{\mathrm{discrete}}$.) Then multiplication by $p$ is bijective, but is not a homeomorphism, since it maps the open subgroup $\mathbf{Z}_p^\mathbf{N}$ onto the non-open subgroup $(p\mathbf{Z}_p)^\mathbf{N}$.

The latter example is not $\sigma$-compact. Actually, every continuous isomorphism of $\sigma$-compact topological groups is a homeomorphism.

However, here's a second countable variant for the question: in the above example, restrict to the subgroup $H$ of $\mathbf{Q}_p^\mathbf{N}$ consisting of elements such that all but finitely many coordinates are in $\mathbf{Z}_p$, and all are in $p^{-1}\mathbf{Z}_p$; this is an open, second-countable subgroup. Then in $H$, multiplication by $p$ maps the whole group $H$ onto a dense proper subgroup of $\mathbf{Z}_p^\mathbf{N}$, which is not even closed.