Why are removable discontinuities even discontinuities at all?

It comes down to what we mean by a function. In specifying the function, we need to specify not just the "rule", but also the domain, i.e., what are you allowed to put into the function?

In particular, to answer your question directly: if we want $f(x)=x+3$ to be defined at $x=2$, then we aren't allowed to say things like $$``f(x)=f(x)\cdot\dfrac{x-2}{x-2},\,''$$ because the expression on the right is not defined at $x=2$.

I think it might be the norm to brush over this kind of thing in introductory calculus courses, because in such a course we tend to assume that the function is defined "everywhere it makes sense" (a frame of mind which is convenient, but carries many problems with it besides just this).

To be a bit more explanatory, if you just say the function $f$ is specified by $$f(x) = \frac{x^{2}+x-6}{x-2},$$ then you haven't specified the function enough to be satisfactory for a really rigorous treatment. In particular, is your function defined at $x=2$, or is it not? Given the way that you have written it down, I would assume that it is not, since, as you point out, we cannot divide by $0$.

The way we deal with this in a rigorous manner is to require the definition of the function to specify the domain. For example, you could say that $f$ has domain $\mathbb{R}$ and is given by $$f(x) = \begin{cases}\dfrac{x^{2}+x-6}{x-2}, & x\neq2, \\ 5, & x=2.\end{cases}$$ Note that the only reason this makes sense is that your function is specified for all $x$ in the domain (in particular, it is clear that there is, for this value of $f(2)$, no discontinuity at all). If we had said "$f$ has domain $\mathbb{R}$ and is given by $$f(x) = \dfrac{x^{2}+x-6}{x-2}$$ for all $x$" then we would run into problems: we claim that $f$ has domain $\mathbb{R}$, but $2\in\mathbb{R}$ and $f(2)$ is not defined.

By contrast, you could specify your function as the function with, for example, domain $[0,1]$ given by $$f(x)=\dfrac{x^{2}+x-6}{x-2},$$ when we see that there is, in a sense, no discontinuity at $x=2$ since $f(2)$ has not been defined. In this case, it makes sense to say $$f(x)=f(x)\cdot\dfrac{x-2}{x-2},$$ since this expression now makes sense for all of the values in the domain of the function. It also now makes sense to make the cancellation and say $f(x)=x+3$, because our knowledge of the domain of the function ensures that we are not dividing by $0$.

What's stopping you from multiplying [a continuous function] by $1$ in the form $\frac{x−2}{x−2}$? Can multiplying by $1$ really introduce a discontinuity to the function?

Ah, but $\frac{x−2}{x−2}$ is not equal to $1$. Or, to be more, precise, it is only equal to $1$ for values of $x$ that are different from $2$. If $x=2$, then $\frac{x−2}{x−2}$ is undefined.

And that's more or less the answer to your question: If you take a given function $f(x)$ that is continuous on all of $\mathbb{R}$, and you multiply it by a function $g(x)$ that is equal to $1$ everywhere except at a single isolated point, at which point $g(x)$ is undefined, then the product $f(x)g(x)$ will be exactly the same as $f(x)$ everywhere except at that point, where it, too, will be undefined.