Define that the order type of an element $x$ in a totally ordered set $X$ is the order type of $\{w \in X\mid w < x\}$. Under this definition, distinct elements of a well-ordered set have distinct order types. Thus, every element of a well-ordered set has a 'canonical name' that uniquely specifies that element, namely its order type.
Anyway, lets call this the 'distinct order type' property, or simply the DOT property. So we'll say that a totally ordered set has the DOT property iff every two distinct elements of that set have distinct order types.
Now clearly, not every totally ordered set has the DOT property. For instance, the elements of $\mathbb{R}$ all have the same order type, so $\mathbb{R}$ certainly doesn't have the DOT property. The same goes for $\mathbb{Z}.$ My question is, do there exist totally ordered sets with the DOT property that are not well-ordered?
There are even densely ordered sets with the DOT property.
Let $\langle X,\le\rangle$ be a linear order, and suppose that there are distinct $x,y\in X$ such that $(\leftarrow,x]\cong(\leftarrow,y]$. Without loss of generality assume that $x<y$, and let $f:(\leftarrow,y]\to(\leftarrow,x]$ be an isomorphism. Clearly
$$h:X\to X:z\mapsto\begin{cases} f(z),&\text{if }z\le y\\ z,&\text{if }y<z \end{cases}$$
is an isomorphism of $X$ into itself. Call a linear order rigid if it admits no non-trivial isomorphism into itself; it follows that every rigid linear order has the DOT property. The following theorem is proved in Brian M. Scott, ‘A characterization of well-orders’, Fundamenta Mathematicæ, Vol. $111$ ($1981$), Nr. $1$, pp. $71$-$76$, freely available here.
Such a linear order, being rigid, has the DOT property, but since it’s dense, it is not a well-order.
Of course $2^{<\omega}=\omega$, so there is a rigid dense linear order of power $2^\omega$. With a little work to handle limit cardinals, the generalized continuum hypothesis implies that there is a rigid dense linear order of every uncountable cardinality.
Added: A nice explicit example of a scattered linear order that has the DOT property but is not a well-order isn’t too hard to construct. Let
$$X=\{\langle n,\alpha\rangle:n\in\omega\text{ and }\alpha\in\omega_n\}\;,$$
and define a linear order $\preceq$ on $X$ as follows: for $\langle m,\alpha\rangle,\langle n,\beta\rangle\in X$, $\langle m,\alpha\rangle\preceq\langle n,\beta\rangle$ iff either $m>n$, or $m=n$ and $\alpha\le\beta$.
Suppose that $\langle m,\alpha\rangle\prec\langle n,\beta\rangle$. If $m=n$, it’s easy to see that $\big(\leftarrow,\langle m,\alpha\rangle\big]$ is not isomorphic to $\big(\leftarrow,\langle n,\beta\rangle\big]$. If $m>n$, the interval $\big[\langle m,\alpha\rangle,\langle n,\beta\rangle\big)$ has order type $\omega_m+\beta$, where $\beta<\omega_n<\omega_m$. Suppose that $\langle k,\gamma\rangle\prec\langle m,\alpha\rangle$, and let $I=\big[\langle k,\gamma\rangle,\langle m,\alpha\rangle\big)$. If $k=m$, the order type of $I$ is less than $\omega_m\le\omega_m+\beta$. If $k>m$, then the order type of $I$ is at least $\omega_{m+1}+\alpha>\omega_m+\beta$. Thus, the order type of $I$ cannot be $\omega_m+\beta$, and $\big(\leftarrow,\langle m,\alpha\rangle\big]$ is not isomorphic to $\big(\leftarrow,\langle n,\beta\rangle\big]$.
Thus, $\langle X,\preceq\rangle$ has the DOT property. A rough sketch of $X$:
$$\cdots\underset{\omega_3}\longrightarrow\underset{\omega_2}\longrightarrow\underset{\omega_1}\longrightarrow\underset{\omega}\longrightarrow$$