What are some ways to show that $1$ is not first-order definable in $(\mathbb{R}, +, 0, \le)$? My question is based on this answer by Asaf Karagila. Emphasis mine.

Indeed as a first-order structure, $(\mathbb{R},≤)$ cannot define the
operations $+,⋅$ or the constants $0,1$. In fact, even as an ordered
group $(\mathbb{R},+,≤)$ cannot define the number 1. These are all simple
exercises that can be given after the second week of a first course
about model theory.

This passage suggests to me that there should be a reasonably direct way to prove that $1$ is not definable in $(\mathbb{R}, +, \le)$ or the similar structure $(\mathbb{R}, +, 0, \le)$.
I'm curious what the canonical way to prove this fact is. I'm also curious if there's a way to prove this that's "syntactic" rather than "semantic" in nature and doesn't appeal to specific models of the theories in question.

Coming up with an informal argument to prove this is relatively straightforward. Let $M^1$ denote the universe of the first model and $M^i$ denote the first model and so on for $(2, ii), (3, iii), (4, iiii), (5, v) \cdots$. Similarly, let $T^i$ be the first theory. For the purposes of this question, I'll assume that the universe of a model is nonempty.
Let $M^i$ be a model of the theory associated with $(\mathbb{R}, +, \le)$. $0$ is the unique solution to the equation with one free variable $\forall w \mathop. x + w = x$.
Let $M^{ii}$ be the model associated with $(\mathbb{R}, +, 0, \le)$.
I can think of two ways to show the non-first-order definability of $1$. I'm pretty sure that these arguments work.
The first is to note that $M^2$ could be $\{0\}$, since without $1$ we are not guaranteed the existence of a positive element. $1$ is not definable because it is not definable in at least one model of $T^{ii}$, therefore it isn't first-order definable.
The second is to look at the standard model of $T^{ii}$. There is more than one way to extend it to a model of the theory associated with $(\mathbb{R}, +, 0, 1, \le)$. If we pick a non-positive element of $M^2$, then it will satisfy all the multiplication-free sentences that $1$ satisfies in the theory associated with $(\mathbb{R}, +, \cdot, 0, 1, \le)$. Since the choice of $1$ is not unique in at least one model, $1$ is not first-order definable.

The intuitive idea behind these arguments, from my perspective, is to locate specific models where there either a) are no suitable candidates for $1$ or b) is more than suitable candidate for $1$.
However, first-order-definability feels like a syntactic property and it feels like I'm using a hypothesis without stating it by looking at specific models and arguing about the reasonableness of extending them with interpretations of new function symbols.
What's the canonical way to prove this fact? Also, is there a way to prove it that's more directly syntactic?
 A: I don't really understand your proposed approach, so I can't comment on it.

I'm not sure it's "canonical" per se, but in my opinion the most natural way to attack a problem like this is via automorphisms.
One of the most important properties of first-order logic is its isomorphism-invariance (and indeed this is usually taken as one of the criteria for being a "logic" in the first place - in particular, everything that follows holds for second-order logic, infinitary logics, etc.):

Suppose $\mathcal{A},\mathcal{B}$ are structures in the same language and $f:\mathcal{A}\rightarrow\mathcal{B}$ is an isomorphism. Then for every formula $\varphi$ in the language of these structures and every tuple $a_1,...,a_n\in\mathcal{A}$ we have $$\mathcal{A}\models\varphi(a_1,...,a_n)\iff \mathcal{B}\models\varphi(f(a_1),...,f(a_n)).$$

We can use this to prove for example that $1$ is not a first-order definable element of the structure $\mathcal{R}=(\mathbb{R};\le,+)$ as follows. Let $f:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto 2x$. This is an automorphism of the structure $\mathcal{R}$ and $f(1)=2$, so by the fact above we get in particular that for each formula $\varphi(x)$ in our language we have $\mathcal{R}\models \varphi(1)\iff\mathcal{R}\models\varphi(2)$. Consequently there is no formula in our language which holds, in $\mathcal{R}$, of exactly the element $1$. And recall that this is what it means for an element to be (un)definable:

Given a structure $\mathcal{A}=(A;...)$ and an element $a\in A$, we say $a$ is (parameter-freely-)definable in $\mathcal{A}$ iff there is some formula $\varphi(x)$ in the language of $\mathcal{A}$ such that $$\{u\in A: \mathcal{A}\models \varphi(u)\}=\{a\}.$$

Note that this is very semantic, contra your request. This is however part of the definition of the term, and it really only ever makes sense to talk about definability relative to a structure, although unfortunately we do often abuse language. See here for a summary of definability (both with and without parameters).

Now the above approach is extremely "coarse," and - unsurprisingly - doesn't always work. For example, the structure $\mathcal{N}=(\mathbb{N};<)$ has no automorphisms at all (= is rigid), and so we can't use the idea above to rule out any definability results at all, but of course only countably many subsets of $\mathbb{N}$ could possibly be definable in $\mathcal{N}$ (and if you want to think about definable elements rather than relations, consider an uncountable rigid structure like the linear order $(\omega_1;<)$).
For more subtle results we need to use more specific properties of the logic in question. For example, the set $E$ of even numbers is not first-order definable in $\mathcal{N}$ but it is second-order definable in $\mathcal{N}$, so if we want to prove the non-first-order definability of $E$ in $\mathcal{N}$ we'll need to use a fact which is more specific to first-order logic. For example, we have the notion of back-and-forth systems, a kind of "isomorphism analogue" for which first-order logic (but not second-order logic!) satisfies an appropriate invariance principle; these are often presented game-theoretically (as Ehrenfeucht-Fraisse games), and I've written a number of arguments here on MSE using this presentation.

The semantic nature of definability notwithstanding, there is a "definability-flavored" question which is purely syntactic. Suppose I have a theory $T$. I can ask whether there is a formula in the language of $T$ which $T$-provably has some property. For example:

Is there a formula $\varphi(x,y,z)$ such that $$\mathsf{Pres}\vdash(\forall x,y\exists !z[\varphi(x,y,z)])\wedge(\forall x,y,x',y',z[\varphi(x,y,z)\wedge\varphi(x',y',z)\rightarrow x=x'\wedge y=y']),$$ where $\mathsf{Pres}$ is Presburger arithmetic?

This essentially asks whether Presburger arithmetic can implement a pairing operation. In case you're curious, the answer is negative (and despite being a syntactic question there's a semantic argument for this!).
We could also make allowances for incompleteness in our theory by permitting some casework (I'll talk about unary relations specifically for simplicity). Given a theory $T$ in a language $\Sigma$ and a sentence $\psi$ in the language $\Sigma\sqcup\{U\}$ for some fresh unary relation symbol $U$, say that $T$ almost defines a $\psi$-solution iff there is a finite sequence $\theta_1(x),...,\theta_n(x)$ of $\Sigma$-formulas such that $$T\vdash\bigvee_{1\le i\le n}\forall x(\psi[\theta_i/U]),$$ where $\psi[\theta_i/U]$ is the $\Sigma$-formula gotten by replacing "$U$" with "$\theta_i$" throughout $\psi$. This turns out to be equivalent to every model of $T$ having a parameter-freely-definable set satisfying $\psi$ in the obvious sense (this uses compactness).
A: In addition to the automorphisms and EF-games that Noah discussed in his answer, another (more syntactic) way to prove non-definability is by understanding all the formulas up to logical equivalence, relative to the complete theory of the structure in question.
Let $T = \mathrm{Th}(\mathbb{R},0,+,-,\leq)$. Note that I'm including $-$ in the language. Sometimes adding definable symbols to the language is necessary to get quantifier elimination. If we want to show that something is not definable, we're free to add more symbols to the language: if it's not definable in the expanded structure, it won't be definable in the reduct. And in this case, $-$ is actually definable in $(\mathbb{R},0,+,\leq)$, by the formula $x+y = 0$, so adding $-$ to the language doesn't even allow us to define any new sets.
Now the theory $\mathsf{ODAG}$ of ordered divisible abelian groups, in the language $\{0,+,-,\leq\}$, is complete and has quantifier elimination [see Corollary 3.1.17 in Marker's Model Theory: An Introduction, for example].
Since $(\mathbb{R},0,+,-,\leq)$ is an ordered divisible abelian group, $T = \mathsf{ODAG}$. So every formula is equivalent, in every model of $T$, to a quantifier-free formula.
What are the possible quantifier-free formulas in one free variable $x$? Every term in $x$ is equivalent to one of the form $nx$, for $n\in \mathbb{Z}$ (where $0x$ is $0$ and for $n>0$, $nx$ is $x+\dots +x$ ($n$ times) and $(-n)x$ is $-(x+\dots +x)$ ($n$ times)). So every atomic formula in $x$ is equivalent to one of the form $nx = mx$ or $nx \leq mx$ for $n,m\in \mathbb{Z}$. Subtracting $mx$ from both sides, every atomic formula in $x$ is equivalent to $kx = 0$ or $kx \leq 0$ for some $k\in \mathbb{Z}$. The first formula defines $\{0\}$ in $\mathbb{R}$, and the second formula defines $\{x\mid x\leq 0\}$ in $\mathbb{R}$. So any Boolean combination of atomic formulas in $x$ defines some union of the sets $\{x\mid x < 0\}$, $\{0\}$, and $\{x\mid 0 < x\}$. This shows that there are only $8$ formulas in one free variable up to logical equivalence modulo $T$. And only one of these $8$ defines a singleton.
