How to come up with this isomorphism?

A side note: this is an exercise. You would expect that whoever made it would make it reasonably easy to solve. In fact, they might have chosen the isomorphism map first (an arbitrary bijection, say, $f(x)=x-1$) and used that to build up the operations ($x\odot y=f(f^{-1}(x)f^{-1}(y))=(x+1)(y+1)-1=xy+x+y$ and similar for $\oplus$). So, with this being an exercise, you can just expect that the rule won't be too complicated. However, the skill you need to have to figure out what the rule is, is universal and applicable to real-life situations. Probably worth a few words about what you could do, in general.

First of all, an isomorphism carries over all algebraic properties of the structure. (All properties expressible via the two operations involved.) It is basically renaming elements but keeping their relationship intact. Thus, if an element in one structure has some sort of unique status, it must map to the element in the other structure with the same status.

Because zero (the neutral for the "addition" operation) is unique, the isomorphism must map the neutral into neutral. So we know $0$ maps to $-1$, and $1$ maps to $0$. However, a lot of other real numbers now have "special" status. For example, $2=1+1$ must map into $0\oplus 0=1$; $3=2+1$ must map into $1\oplus 0=2$ etc. - every natural number $n$ maps to $n-1$.

At this point you may already have a good idea that every real number $x$ maps to $x-1$, but let's say you still don't see the map very clearly. Can we see mapping of other numbers? For example, if $n\in\mathbb N$, $-n$ is the unique inverse of $n$ for $+$, so it needs to map into the unique inverse of $n-1$ for $\oplus$. What is it? Solve for unknown $m$: $(n-1)\oplus m=-1$, i.e. $n+m=-1$ i.e. $m=-n-1$ and so negative numbers $-n$ are also mapped into $-n-1$.

What about fractions? First, $1/2$ is the unique inverse of $2$ for $\cdot$, so it needs to map into the unique inverse of $1$ for $\odot$. More generally, $1/n$ maps to the unique inverse of $n-1$ for $\odot$. Again, solve $(n-1)\odot q=0$ i.e. $(n-1)q+n-1+q=0$ for $q$, and you will get $qn+n=1$ i.e. $q=1/n-1$. So the fraction $1/n$ maps to $1/n-1$.

And so on... I will leave to you to prove that any fraction $p/q$ must map into $p/q-1$. This should be already enough to have a good guess that every $x\in\mathbb R$ maps into $x-1$. If you still cannot see it, you can go even further: in $\mathbb R$, the relation $\le$ is an algebraic property becuse $x\le y\iff (\exists z)(y=x+z^2)$. So let's suppose $x\le y$, so there is $z$ such that $y=x+z^2$. Applying $f$, you conclude that $f(y)=f(x)\oplus(f(z)\odot f(z))$. Let's now see what it means:

$$\begin{array}{rcl}f(y)=f(x)\oplus (f(z)\odot f(z))&\iff&f(y)=f(x)\oplus (f(z)^2+2f(z))\\&\iff&f(y)=f(x)+f(z)^2+2f(z)+1\\&\iff&f(y)=f(x)+(f(z)+1)^2\end{array}$$

which tells you that $f(x)\le f(y)$. The conclusion is: the isomorphism must also preserve the ordering. Thus, if you have proven that $x$ maps to $x-1$ for all rational numbers, simply because irrational numbers are uniquely defined by which rational numbers they are wedged in between, you get that the same formula is valid for all real numbers.

To summarise: all we did to reconstruct the isomorphism was to spot the unique role the original elements of $\mathbb R$ play with respect to the operations $+$ and $\cdot$, and calculate which elements play the same role with respect to the new operations $\oplus$ and $\odot$. In this problem it is easy, because of the second observation: every number in $\mathbb R$ plays a unique role. Thus the isomorphism can be reconstructed fully. (Probably it is an important point for you to try to fully understand how ordering happens to be an algebraic property on $\mathbb R$.) In other cases, you may have ambiguities - sometimes there will be more than one element that satisfies some property, which may give rise to multiple isomorphisms.

As I hinted in your link, any set bijection $\,h\,:\,\tilde R\to R\,$ serves to transport the ring structure of $\,(R,+,\cdot,0,1),$ to $\,(\tilde R,\oplus,\odot,\tilde 0,\tilde 1) \,$ by defining operations in $\,\tilde R\,$ so $\,h\,$ is a ring isomorphism, i.e.

$$\begin{align} h\,\text{ is a ring hom}\!\iff h(a \oplus b)\, &\ \ \ = \ \ \ \ \ \ h(a) + h(b),\quad h(\tilde 0) = 0\\ h(a \odot b)\, &\ \ \ = \ \ \ \ \ \ h(a)\ \cdot\ h(b),\quad\, h(\tilde 1) = 1\\[.3em] \iff\ \ \ \ a \oplus b\ \ &=\ \color{#c00}{h^{-1}}(h(a) + h(b)),\quad \tilde 0 = h^{-1}(0)\\ a \odot b\ \ &=\, \color{#0a0}{h^{-1}}(h(a)\ \cdot\ h(b)),\quad\, \tilde 1 = {h^{-1}}(1)\\[.3em] {\rm when}\,\ h(x) = x\!+\!1\!:\,\ \ a\oplus b\ \ &= \color{#c00}{\!-\!1+}(a\!+\!1)+(b\!+\!1) \,=\, a\!+\!b\!+\!1\\ \,\ a\odot b\ \ &= \color{#0a0}{\!-\!1+}(a\!+\!1)\ \cdot\ (b\!+\!1) \,=\, a\cdot b\!+\!a\!+\!b \end{align}\qquad \qquad\qquad\qquad$$

which yields the sought operations on the transported ring structure $\,\tilde R\,$ (note $\,\color{#c00}{h^{-1}}(x) = \color{#c00}{-1+}x).\,$

In cases like this it is usually easy to derive the bijection $h$ from the given transported ring operations, e.g. let $\,g = h^{-1}.\,$ You have already correctly deduced that $\,\color{#c00}{g(1)=0},\,$ hence $\, g(n\!+\!1) = g(n)\oplus g(1) = g(n)\!+\!\color{#c00}{g(1)}\!+\!1 = g(n)\!+\!1.\,$ This linear recurrence $\,g(n\!+\!1) = g(n)\!+\!1\,$ has solution $\,g(n) = \color{#c00}{g(1)}\!+\!n\!-\!1 = n\!-\!1,\,$ thus $\,h(n) = g^{-1}(n) = n\!+\!1,\,$ and, as above, trivial algebra verifies that this bijection $\,h\,$ yields the given transported operations.

To gain further intuition, let's consider some other common simple transported ring structures.

If $R$ is countable then there is a bijection $\,h\,:\,\Bbb N\to R,\,$ so we can choose $\tilde R$ to be $\Bbb N$ and then we can think of the naturals as index numbers for the elements of $R$ (e.g. computer memory addresses of their data structure). Then $h$ maps the index to the ring element, and its inverse $h^{-1}$ maps the ring element to its index. To perform a multiplication on indices $\,a\odot b\,$ we dereference them to their ring elements $\,h(a),h(b)\,$ then do the ring multiplication $\,h(a)\cdot h(b),\,$ then return the index of the product $\,h^{-1}(h(a)\cdot h(b)).$ Your example works the same way except it indexes (labels) each ring element $\,a\,$ not by a natural but instead by another ring element, viz. $\,a-1.\,$

A more common example of transported ring structures are rings of normal forms, e.g. consider $\,R = \Bbb Z/m = $ ring of integers $\!\bmod n,\,$ whose elements are cosets $\,[a] = a+n\Bbb Z\,$ with operations $[a]+[b] = [a+b],\ $ $\, [a]\cdot [b] = [a\cdot b].\,$ As usual, a convenient label (index) for the coset $[a]$ is its least nonnegative element $\,a\bmod n,\,$ so our index set is $\,\tilde R = \{0,1,2,\ldots, n\!-\!1\}$, and our maps are $\,h(k) = [k] = k+n\Bbb Z\,$ and index map $\,h^{-1}[k] = k\bmod n.\,$ As for transported operations,

$$\begin{align}\text{here we have that }\, &h(a)\cdot h(b) = [a]\cdot [b] = [a\cdot b]\\[.3em] \text{thus }\ a\odot b = h^{-1}(&h(a)\cdot h(b)) = h^{-1}[a\cdot b] = (a\cdot b)\bmod n \end{align}\qquad\qquad$$

i.e. $\ a\odot n = (a\cdot b)\bmod n.\,$ Similarly $\, a\oplus b = (a+b)\bmod n\,$ so we obtain the well-known representation of $\,\Bbb Z/n\,$ using least natural rep's in $\{0,1,2,\ldots,n-1\},\,$ with modular arithmetic operations, obtained by pulling back the ring operations along the normal-form map.