I would define “rigorous” as follows:
A piece of mathematics (for example, a definition, statement of a proposition, or proof) is said to be rigorous when an individual with an appropriate amount of mathematical training could take this piece of mathematics and translate it into a reasonable formal logical system.
This definition, by its nature, is not rigorous (as defined above). In fact, it is somewhat ambiguous, even ignoring that it is impossible to translate it into a formal system.
Here is an example of a rigorous definition:
A set is said to be a subsingleton when any two elements of the set are equal to each other.
This corresponds to the definitional extension of some sort of set theory (perhaps ZFC) given by $subsingleton(x) :\equiv \forall y \forall z (z \in x \land y \in x \to z = y)$. A trained mathematician would easily be able to translate the definition into this formal definitional extension. Therefore, the definition is rigorous.
An example of a nonrigorous definition would be
A function $f : \mathbb{R} \to \mathbb{R}$ is continuous when you can draw the graph of $f$ with a pencil without lifting the pencil off the paper.
While this definition is intuitive, it gives no clue for how to translate the notion into a formal mathematical language.
An example of a rigorous proof would be the following:
Theorem: Let $G$ be a group and consider three elements $x, y, z$ of the group. Suppose $x \cdot y = x \cdot z$. Then $y = z$.
Proof: we have $y = e \cdot y = (x^{-1} \cdot x) \cdot y = x^{-1} \cdot (x \cdot y) = x^{-1} \cdot (x \cdot z) = (x^{-1} \cdot x) \cdot z = e \cdot z = z$ $\square$
An appropriately trained mathematician would easily be able to turn the above proof into a formal proof in, say, the first-order theory of groups, using the various axioms of a group. Thus, the above proof is a rigorous one.
An example of a non-rigorous proof would be as follows:
Theorem: any set $S \subseteq \mathbb{N}$ which isn’t empty has a least element.
Proof: let’s suppose $S$ doesn’t have a least element. We can start with an element $s_0 \in S$, then pick some $s_1 < s_0$ such that $s_1 \in S$, then pick $s_2, s_3$, and so on. This gives us an infinite descending sequence of natural numbers. But that is absurd, since natural numbers just don’t work that way. $\square$
This is not a rigorous proof if we are working within the usual realm of mathematics. The part about picking $s_0, s_1$, and so on is perfectly rigorous if we assume the axiom of dependent choice. But what is not rigorous is the hand-wavy argument that we can’t have an infinite descending sequence of natural numbers. We would need to appeal to some axiom or to the definition of “natural number” to draw this conclusion.
But note that many ancient Greeks did essentially take as an axiom that an infinite descending sequence of natural numbers is impossible; if you follow in their footsteps, the proof could be considered rigorous.
The moral of the story is that rigor is, in part, a social construct, but in part rooted in formal logic. Mathematicians have a complex system of conventions for what natural language means, how to phrase logical arguments, what principles or theorems should be considered common knowledge, which steps are simple enough that they can be skipped, and more. Evaluating whether a given proof is rigorous requires knowing the entire context of the proof and is, to some degree, unavoidably subjective.
