There are many bijective functions that map $\mathbb{R}$ to $(-1,1)$, in particular: $$f\left(x\right)=\frac{e^{2x}-1}{e^{2x}+1}$$
(Of course there are others, such as $g\left(x\right)=\frac{2}{\pi}\arctan(x)$, etc., but I like mine since $f\left(0\right)=0$ and $f^\prime\left(0\right)=1$.)
$f\left(x\right)$ is indeed a bijection, and it also preserves order, since $x_1\le x_2$ implies $f\left(x_1\right)\le f\left(x_2\right)$. I'll omit the proofs for the sake of wordiness. Incidentally, my conjecture is that all order-isomorphic functions that map $\mathbb{R}$ to $(-1,1)$ are sigmoid functions, but I digress.
My first question is about the nature of the isomorphism $f$. Obviously $f$ is order-preserving, but is it operation-preserving? Does there exist an operation $\#$ on $(-1,1)$ such that $f(x_1)\#f(x_2)=f(x_1+x_2)$? or $\ast$ such that $f(x_1)\ast f(x_2)=f(x_1\cdot x_2)$?
Also, my second question: In layman's terms, two sets that are "isomorphic" are "essentially the same." That is, you could treat both sets as the same set, except that maybe the elements have different names*. I don't see how this is possible, since $(-1,1)$ has a least upper bound, whereas $\mathbb{R}$ does not have a least upper bound. $\mathbb{R}$ and $(-1,1)$ don't "look" the same. All the other isomorphic sets I can think of "look" (structurally) the same.
*The sets $\left\{\text{one},\text{two},\text{three}\right\}$ and $\left\{\textit{uno},\textit{dos},\textit{tres}\right\}$ are order- and operation-isomorphic.
EDIT: I just discovered that the function $f(x)$ is exactly $\tanh(x)$.
You just defined $\#$ and $*$ in terms of $f$. Namely, the operations: $$a\# b=f\left(f^{-1}(a)+f^{-1}(b)\right)\\a*b=f\left(f^{-1}(a)\cdot f^{-1}(b)\right)$$
Obviously $f$ respects these operations, and it's not easy to see that they are unique. Furthermore $(0,1)$ with $\#$ and $*$ forms a field, which is isomorphic to $\mathbb R$ via $f$. This is also known as transport of structure.
As for your second question, note that internally $(0,1)$ does not have an upper bound. If two structures are isomorphic this means that you can treat them as the same structure, up to replacing the elements. So only things which are internally definable from this structure should be relevant.
Note that as topological spaces $(0,1)$ and $\mathbb R$ are the same, but as metric spaces they are not isomorphic because $(0,1)$ has a finite width and $\mathbb R$ not. You can't use everything you know about the particular set, but rather the things which are preserved by the isomorphism.