What I understand :
A group is characterized by how any two of its elements interact. It is the "structure" of the group that describes it. Thus two groups having the same structure are basically the same (isomorphic).
The additive group $\Bbb R$ can be regarded as the set of all sliding symmetries of the "number" line. Any number can be seen as the sliding action that takes $0$ to that number.
Similarly the multiplicative group $\Bbb R^+$ is the set of all stretching/squishing symmetries of the number line. A number here represents the action that takes $1$ to that number.
The problem :
It seems to me that the two groups mentioned above are very different in the sense that sliding and stretching/squishing are very different types of actions. But, the two groups are isomorphic. I would always see two isomorphic groups as being the same; however, after looking at $\Bbb R$ and $\Bbb R^+$ as groups of symmetries of the number line, I'm not so sure I would consider them as being the same.
P.S. I am not familiar with the concept of a "group action"
Best Answer
Everything I talk about can be formally defined, but as your question seems to be one on intuition I will leave everything as intuition.
Two groups $G$ and $H$ are isomorphic if the have the same group structure - that is we can relabel elements of $G$ and the group operation to get $H$, in your example we replace '$0$' with '$1$' and $x$ with $e^x$ say to go from $\mathbb{R}$ to $\mathbb{R}^+$ (and obviously $+$ with $\times$).
But the structure you refer to in the question is the action of $G$. Two actions of a group need not be the same in structure (despite $G$ obviously being isomorphic to itself). For example $C_2\times C_2$ can be thought of as the subgroup of $S_4$ generated by $(1,2)$ and $(3,4)$ or the subgroup generated by $(1,2)(3,4)$ and $(1,3)(2,4)$. These actions are not equivalent (one is transitive, the other isn't). It so happens that the two actions you give are equivalent, by the same relabelling that gives the isomorphism.