First of all, there are two aspects to Lipschitz maps between metric spaces: Local (having implications such as differentiability properties, etc) and global. The two aspects have different non-metric formalizations. Below, I will be mostly addressing the local aspect. If you are interested in the global aspect (appearing under the umbrella of "Coarse Geometry"), read section 2.1 of
Roe, John, Lectures on coarse geometry, University Lecture Series 31. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-3332-4/pbk). vii, 175 p. (2003). ZBL1042.53027.
The key difference is that instead of uniformities in my answer Roe uses "coarse structures" ("coarsivities"). The definition is almost the same but instead of been closed under taking supersets, coarsivities are closed under taking subsets. Large-scale Lipschitz maps are then defined as "bornologous maps," see also here.
There are some interesting relationships between micro and macro aspects of Lipschitz maps, but I will not go into this.
One can define filtered uniformities for more general posets than ${\mathbb Z}$ and non-Hausdorff uniformities, which will lead to the notion of Lipschitz maps between nonmetrizable spaces. I will not do this.
Definition 1. A Hausdorff uniformity ${\mathcal U}$ on a set $X$ will be called ${\mathbb Z}$-filtered if it comes equipped with a map ${\mathbb Z}\times {\mathcal U}\to {\mathcal U}, (a,U)\mapsto U_a$, such that
$$
a\le b\Rightarrow U_a\subset U_b,
$$
and $U_0=U$ for all $U\in {\mathcal U}$.
For some $U\in{\mathcal U}$, the subset $<U>=\{U_n: n\in {\mathbb Z}\}$ is a basis of ${\mathcal U}$. (I will refer to such $U$'s as "generators" of ${\mathcal U}$.)
Example. Suppose that ${\mathcal U}={\mathcal U}_d$ is the metric uniformity defined by a metric $d$ on $X$: A basis ${\mathcal B}$ of ${\mathcal U}_d$ is given by subsets of the form $U(\epsilon)=\{(x,y)\in X^2: d(x,y)<\epsilon\}$, $0<\epsilon<\infty$. Then ${\mathcal U}$ consists of all subsets of $X^2$ containing elements of ${\mathcal B}$. For $n\in {\mathbb Z}$, $U=U(\epsilon)$, define
$$
U_n:= \{(x,y)\in X^2: d(x,y)< 2^n \epsilon\}.
$$
Thus, for each $U\in {\mathcal B}$, $<U>$ is a basis of ${\mathcal U}$.
For each $U\in {\mathcal U}$ which is not in ${\mathcal B}$, I will take
$$
U_n:= U, n\in {\mathbb Z}.
$$
(I am sure somebody can come up with a more imaginative choice here.)
In what follows, I will simply say that a uniformity is filtered rather than ${\mathbb Z}$-filtered. Thus, every metric uniformity is filtered.
Lemma 1. Each filtered uniformity ${\mathcal U}$ is metrizable.
Proof. The proof is quite standard. Pick a generator $U$ of ${\mathcal U}$. For each pair $x, y\in X$ define $\phi_U(x,y)$ as
$$
\inf \{2^n: n\in {\mathbb Z}, (x,y)\in U_n\}.
$$
Then set
$$
d_U(x,y)=\inf \sum_i \phi_U(x_i, x_{i+1})
$$
where the infimum is taken over all chains $x=x_1, x_2...,x_{k-1}, x_k=y$. Then $d_U$ is a metric (recall that ${\mathcal U}$ is assumed to be Hausdorff) and its uniformity is equivalent to ${\mathcal U}$. qed
I will use the notation $d_U$ for the metric on $X$ given by this construction.
Suppose that ${\mathcal U}, {\mathcal V}$ are filtered uniformities on sets $X, Y$ and $f: X\to Y$ is a map. Then $f$ induces the map $f: X^2\to Y^2$,
$$
f(x_1, x_2)= (f(x_1), f(x_2)).
$$
Definition 2. A map $f: X\to Y$ (between two sets equipped with filtered uniformities) is said to be Lipschitz (in the sense of uniformities) if for each generator $V\in {\mathcal V}$, there exists a generator $U\in {\mathcal U}$ such that for all sufficiently small $n\in {\mathbb Z}$,
$$
f(U_n)\subset V_n.
$$
Equivalently, one can require this property for all generators $V$; also, equivalently, one can require this for all sufficiently small generators $U$.
Definition 3. Two uniformities ${\mathcal U}, {\mathcal V}$ on the same set $X$ are Lipschitz-equivalent (in the sense of uniformities) iff the identity map $(X, {\mathcal U})\to (X, {\mathcal V})$ is bi-Lipschitz (i.e. is Lipschitz and its inverse is also Lipschitz). A bi-Lipschitz structure on a set $X$ is the equivalence class of a filtered uniformity on $X$.
I will omit the proofs of the lemmata below:
Lemma 2. If ${\mathcal U}, {\mathcal V}$ are metric uniformities on metric spaces $X, Y$, a map $f: X\to Y$ is Lipschitz if and only if it is a Lipschitz map in the sense of Definition 2.
Lemma 3. If ${\mathcal U}$ is a filtered uniformity, then for any two generators $U, V$ of ${\mathcal U}$ the metrics $d_U, d_V$ are Lipschitz-equivalent.
Lemma 4. Suppose that $(X,d)$ is a metric space, ${\mathcal U}={\mathcal U}_d$ is the corresponding metric uniformity with a generator $U$. Then the metrics $d, d_U$ are Lipschitz-equivalent in the traditional sense.
So, here you have it: A non-metric definition of bi-Lipschitz equivalence and structure.
Old answer:
Since nobody came up with anything better, let me expand my comment here, at the MSE version of the question.
The substantive part of your question is to give a "metric-free" definition of the Lipschitz equivalence class of metrics on a given topological space. I will deal only with the case of compact spaces since the answer is much cleaner in this setting.
To every compact metric space $(X,d)$ one associates its Lipschitz algebra which is the Banach *-algebra of complex-valued Lispchitz functions $Lip(X)$ on $X$, equipped with the Lipschitz norm
$$
||f||_{Lip}= ||f|| + \sup_{x\ne y} \frac{||f(x)-f(y)||}{d(x,y)}.
$$
Here $||f||$ is the supremum-norm of $f$.
To the best of my knowledge, these algebra were first analyzed in detail by Donald Scherbert in
Banach algebras of Lipschitz functions, Pacific J. Math. 13 (1963) 1387--1399.
Scherbert proves results for general metric spaces; here are
are some basic results that he proved adopted to the compact case:
$Lip(X)$ is a (unital, which I will assume throughout) *-Banach algebra, i.e. a topological *-algebra (up to an isomorphism in the category of topological *-algebras) which admits a Banach norm. (I am sure somebody who knows analysis better than I do can come up with a reference to a "metrization theorem" for topological *-algebras. Frechet is a necessary condition.)
Every *-Banach algebra is isomorphic to $Lip(X)$ for some compact metric space $X$.
Metrics $d_1, d_2$ on $X$ are (boundedly) equivalent (i.e. the identity map is bi-Lipschitz) if and only if $Lip(X,d_1)=Lip(X,d_2)$ as subsets of $C(X)$.
Every morphism $Lip(X_1,d_1)\to Lip(X_2,d_2)$ is induced by a Lipschitz map $(X_1,d_1)\to (X_2,d_2)$. In particular,
two metric spaces $(X_1,d_1), (X_2,d_2)$ are bi-Lipschitz homeomorphic if and only if
$Lip(X_1,d_1)\cong Lip(X_2,d_2)$.
Thus, Scherbert's results yield a "metric-free" description of the Lipschitz category, i.e. the category of compact metric spaces where morphisms are Lipschitz-continuous. This does not quite answer your question which, in the language of Banach algebras can be formulated as follows:
Given a compact metrizable space $X$, describe subalgebras $A\subset C(X)$ (the $C^*$-algebra of complex-valued continuous functions on $X$) such that there exists a metric $d$ on $X$ for which $A=Lip(X,d)$.
One can give several easy necessary conditions:
(1) $A$ is "regular" (see Scherbert's paper).
(2) $A$ is "large" in the sense that every maximal ideal in $C(X)$ is generated by an element of $A$.
(3) $A$ is "small" in the sense that $C(X)$ and $A$ admit Banach norms $||\cdot||$ and $||\cdot||_A$ making them, respectively, a $C^*$-algebra and a Banach *-algebra, such that the inclusion map
$$
(A, ||\cdot||_A)\to (C(X), ||\cdot ||)
$$
has unit norm and is compact. (The latter is the Arzela-Ascoli theorem.)
I suspect that a complete list of properties characterizing Lipschitz subalgebras of $C^*$-algebras is well-known (to the right group of people).
