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"
Let me add a bit of spin to the existing answers: what's going on here is that you find group actions more natural things than groups "on their own." This is totally reasonable and personally I agree, but the point is that the notion of group "on its own" is specifically intended to abstract away from the specific action. Why do we want to do this? Well, one answer is that groups, rather than group actions, are also interesting in their own right; a less "pure" answer would be that - as often happens - studying the more "abstract" objects can yield data in the "concrete" case.
However, at the same time it's worth noting that there is also a notion of isomorphism of group actions: if I have an action $\alpha$ of a group$G$ on a set $X$ and an action $\beta$ of a group $H$ on a set $Y$, then $\alpha$ and $\beta$ are isomorphic if there are bijections $f:G\rightarrow H$ and $g:X\rightarrow Y$ which "make everything commute," that is, such that for all $c\in G$ and $x,z\in X$, we have $\alpha(c,x)=z\iff \beta(f(c),g(x))=g(z)$. With this notion the actions of $\mathbb{R}$ on $\mathbb{R}$ and $\mathbb{R}^+$ on $\mathbb{R}$ you describe are not isomorphic! This is a good exercise; as a hint, think about fixed points (can an action of the latter ever move $0$?).
That said, plenty of actions we may think of as not being the same are in fact isomorphic; e.g. the mutliplicative of $\mathbb{R}^+$ on $\mathbb{R}^+$ is isomorphic in the above sense to the additive action of $\mathbb{R}$ on $\mathbb{R}$. So there is still some potential non-intuitiveness here.
The point is that there are two distinctions you need to have in mind: equality versus isomorphism and groups versus group actions. Once you internalize these, I think the confusion will disappear.