Nice try, mainly I meant nice try in disproving yourself, because when testing the waters, the ability to prove/disprove something based on how you decided to define things is a great way of determining if your definitions make any sense mathematically.
However, instead of dividing by zero, which happens to be impossible, a better problem would be to divide by something small. So small that it is almost $0$, but not quite there yet. For lack of better term, we'll call this magical almost $0$ the number $\epsilon$. Funny thing is we can divide by $\epsilon$, especially if the thing we are dividing by is almost $0$ too.
I want to look at the following function:
$$f(x)=\frac{x^2-1}{x-1}$$
Particularly, I want to study it at $x=1$, though this results in division by $0$. So instead, I study the function at $x=1+\epsilon$ and $x=1-\epsilon$.
$$\begin{array}{c|c|c}\epsilon&f(1+\epsilon)&f(1-\epsilon)\\\hline1&3&1\\0.1&2.1&1.9\\0.01&2.01&1.99\\0.001&2.001&1.999\\\vdots&\vdots&\vdots\\\epsilon&2.000\dots&1.999\dots\end{array}$$
So I guess we can agree that it makes the most sense that $f(1\pm\epsilon)=2!$ This... idea. It is known as a limit, and it usually only works when we have:
$$f(x)=\frac00$$
In the event that we have $\frac10$, the numbers will get infinitely big. Try $f(x)=\frac1x$ around $x=0$ for example.
A good note is that we avoid any problems with this idea of division by zero since we aren't actually using $0$. We're just using very small numbers, where all basic math still holds. This way, we can't construct weird proofs of $2=0$.