Can this be considered a "rigorous" definition of a limit?

No, this is not equivalent to the usual definition of a limit. Consider for example $$f(x) = \begin{cases} x \sin(1/x) & x \neq 0 \\ 0 & x = 0 \end{cases}$$

Then $\lim_{x \to 0} f(x) = 0$ as can be easily checked (note that $|f(x)| \le |x|$ for all $x \in \mathbb{R}$).

However with your definition then $f$ wouldn't have this limit. Indeed, applied to $a = 0$ and $l = 0$, this would mean that we would have $$|x| < |y| \iff |x \sin(1/x)| < |y \sin(1/y)|.$$ This is false. Let $x = \frac{2}{3\pi}$ and $y = \frac{1}{\pi}$. Then $|x| < |y|$, however $|f(x)| = \frac{2}{3\pi} > |f(y)| = 0$.

I guess what you were going for was "the closer $x$ is to $a$, then the closer $f(x)$ is to $l$"? The problem with that, as you can see, is that $f(x)$ can oscillate a lot around $l$ even though it gets closer and closer to it.

This definition says "whenever $x$ gets closer to $a$, $f(x)$ gets closer to $l$". But this is not the usual definition of limit. For example, let's consider $f(x)=x^2$, and $a=0$. Then, since $$\lim_{x \to a} f(x)=0^2=0$$ it is obvious that whenever $x$ gets closer to $0$, then $f(x)$ gets closer to $0$. However, it gets closer to every other negative number as well.

So, according to your definition of limit, it would be that every negative number $l <0$ satisfies $$\lim_{x \to 0}x^2=l$$ which is not the usual notion of limit.

A more severe problem with this definition is that the $\iff$ in the inequality implies that $f$ takes level sets of distance to $a$, to level sets of $f$, so that (on intervals) the only continuous functions according to this concept of limit are linear.

To formalize the property "$x$ closer to $a$ implies $f(x)$ closer to $L$", the arrow in the definition should be only in the forward direction, $\implies$. Although that is a stronger property than existence of a limit, it is also true that this property holds in most cases where there is a limit.