Reduction from Vertex Cover to Dominating Set
If I understand correctly you only have a problem when the graph $G = (V,E)$ of the vertex-cover instance has isolated vertices. In this case you can transform $G$ into a related graph $G' = (V \cup \{x,y \}, E')$ by adding a two new vertices $x$ and $y$  such that  $x$ and $y$ are connected to each other by an edge, and there is an edge between $x$ and each other vertex in $V$. Formally: $E' = E \cup \{ (x,v) \mid v \in V \cup \{y\}\}$.

If $G$ admits a vertex-cover $C$ of size at most $k$, then $G'$ admits a vertex-cover of size at most $k+1$, namely $C \cup \{ x \}$.

If $G'$ admits a vertex-cover $C$ of size at most $k+1$, then $G$ admits a vertex-cover of size at most $k$. This can be easily seen by noticing that $C \setminus \{ x, y\}$ must cover all the edges in $E$, and that $(x,y) \in E'$ ensures that at least one of $x$ and $y$ is in $C$, i.e., $|C \setminus \{ x, y\}| = |C| - |C \cap \{x,y\}| \le |C| - 1 \le k$.

Since $G'$ has no isolated vertex, you can now safely transform it to the graph $H$ of the dominating-set instance (using the known reduction). 
In this way you show that $G$ has a vertex-cover of size at most $k$ $\iff$ $H$ has a dominating set of size at most $k+1$.