Maybe it's a bit of a stretch, but I am surprised this operation, which opens up a whole new world, has not been mentioned yet:
$$\int (\cdot) dx$$
Let me expand a bit. I will more or less copy-paste the same idea three times to highlight the similarity in ideas.
When you play with $\mathbb N_0$, you notice that a) you can add and multiply, b) there is a mildly interesting "successor" operator $S: \mathbb N_0 \rightarrow \mathbb N_0$ with $S(0)=1, S(3)=4$, $S(17) =18$ etc. You notice that almost all numbers have predecessors (are in the image of $S$), but $0$ is not. So you just $$\text{invent a new thing, called } -1,$$ such that $S(-1)=0$. It feels a bit like cheating. You wonder if this $-1$ also needs a predecessor. However, since also you want to compatibly extend your rules of addition and multiplication, you'll notice that you'll need $-2 := 2(-1), -3:= 3(-1)$ etc. anyway, and it turns out that if you define that $-2$ as predecessor of $-1$, $-3$ as predecessor of $-2$ etc., indeed, suddenly you have constructed a new structure which is a) still closed under addition and multiplication, and b) now your operator $S$ is surjective. You have just invented $\mathbb Z$.
When you play with $\mathbb Z$, you notice that a) you can add, subtract, and multiply, and b) for each $k \in \mathbb Z$, there is a mildly interesting "scaling" operator $M_k: \mathbb Z \rightarrow \mathbb Z$ which is linear and satisfies $M_k(1)=k$. You notice that for most $k$, some numbers are in the image of $M_k$, but many, including $1$, are not. So you just $$\text{invent a new thing, called } 1/k,$$ such that $M_k(1/k)=1$. It feels a bit like cheating. You wonder whether $1/k$ also needs a new preimage under $M_k$. However, since also you want to compatibly extend your rules of addition, subtraction, and multiplication, you'll notice that you'll need $(1/k)^2$, $(1/k)^3$ etc. anyway, which give more preimages under $M_k$. It turns out you have constructed a new structure which is a) still closed under addition, subtraction, and multiplication, and b) now your operator $M_k$ is surjective. You have just invented the localisation $\mathbb Z_{(k)} = \mathbb Z [1/k]$. If you do this for all $M_k$ (all prime $k$ suffice), you have invented $\mathbb Q$.
When you play with $\mathbb R$, you notice that a) you can add, subtract, and multiply, and b) there is a mildly interesting "square" operator $\square: \mathbb R \rightarrow \mathbb R$ with $\square(-2.5)= 6.25, \square(\pi)=\pi^2$, $\square(17) =289$ etc. You notice that many numbers have square roots (are in the image of $\square$), but many, including $-1$, are not. So you just $$\text{invent a new thing, called } i,$$ such that $\square(i)=-1$. It feels a bit like cheating. You wonder whether $i$ also needs a square root. However, since also you want to compatibly extend your rules of addition, subtraction, and multiplication, you'll notice that you'll need $3+4i$, $-\pi-7.43i$, $\frac12\sqrt2(1+i)$ etc. anyway, and it turns out that indeed, suddenly you have constructed a new structure which is a) still closed under addition, subtraction, and multiplication, and b) now your operator $\square$ is surjective. You have just invented $\mathbb C$.
When you play with $\mathbb C$, you notice that a) you can add, subtract, and multiply, and b) there is a "differentiation" operator $D: \mathbb C \rightarrow \mathbb C$ which is linear and satisfies $D(ab) = aD(b)+bD(a)$. You notice that to be honest, this $D$ is rather dull as it just sends every number to $0$, i.e. any nonzero number does not have a preimage under $D$. So you just $$\text{invent a new thing, called } x,$$ such that $D(x)=1$. It feels a bit like cheating. You wonder whether $x$ also needs a new preimage under $D$. However, since also you want to compatibly extend your rules of addition, subtraction, and multiplication, you'll notice that you'll need $x^2$, $x^3$ etc. anyway, and you find that $D(\frac12 x^2)=x, D(\frac13 x^3)=x^2$ etc. It turns out you have constructed a new structure which is a) still closed under addition, subtraction, and multiplication, and b) now your operator $D$ is surjective. You have just invented the polynomial ring $\mathbb C[x]$. And if you also want to extend division, you will invent the field of rational functions $\mathbb C(x)$.
So I think this is at least an interesting additional perspective: that in each case we interpret the idea of the extension as making a certain operator surjective by "inventing" preimages for it. Certainly the operator $D$ feels a bit forced, as it is only obvious in hindsight that what we want to find preimages for is not just "the map that sends everything to $0$" but "the map which happens to send everything so far to $0$ because it satisfies the product rule of derivations." To further highlight the analogy, I should have e.g. defined the operator $S$ as "a map that satisfies $S(a)S(b) = S(ab)+a+b$ (which just happens to necessarily be the succesor function $S(n)=n+1$ on $\mathbb Z$).
The other thing that I glossed over here is the technical details of what it means to "compatibly" extend in each case.
As a last interesting remark, the construction of the polynomial ring as stated works over any characteristic $0$ field in place of $\mathbb C$; but if you do that over e.g. $\mathbb Z$, you'll find that you have to construct more than just the polynomials (because e.g. $x$ still does not have a preimage under $D$ in $\mathbb Z[x]$), and this time you more or less invent a divided power structure.