I feel a bit funny posting this as an answer to someone who has written a book with "finite groups of Lie type" in the title.
But I also find the matter both confusing and interesting (and full of conflicting terminology!), so here's my outside take on this, being a topologist of training:
(My own literature references include the standard sources: Carter, Gorenstein-Lyons-Solomon3..., as well as the recent (recommendable!) book by Malle-Testerman):
I thought the most general definition of a finite group of Lie type was close to your definition $\cal L_1$:
Definition ($\cal L_3$): A finite group of Lie type $G$ is a finite group obtained as ${\mathbf G}^F$, where $\mathbf G$ is a connected reductive algebraic group over an algebraically closed field of characteristic $p$, and $F$ is a Steinberg endomorphism (aka "twisted Frobenius", a composite of a Frobenius map and an automorphism of $\mathbf G$), or a group obtained from such a group ${\mathbf G}^F$ by modding out by a central subgroup, or passing to a normal subgroup with abelian quotient group.
Groups of this form will have many "Lie-like" properties, and in particular admit a canonical BN-pair structure. (Admitting a BN-pair structure would be another possible definition!)
Some authors restrict $\cal L_3$ by assuming $\mathbf G$ is semi-simple, or even simple, and some authors don't allow central quotients or passing to subgroups contained in the commutator subgroup. (One reason to allow the central quotients and subgroups is that otherwise the construction rarely produces simple groups: $|{\mathbf G}^F|$ only depends on the isogeny class of $\mathbf G$, e.g., adjoint and simply connected form have the same order, so ${\mathbf G}^F$ itself will never be simple unless $\mathbf G$ has associated Cartan matrix of determinant $1$.) Sticking with reductive (rather than simple or semi-simple) has the advantage that Levi subgroups in finite groups of Lie type are still finite groups of Lie type, which is useful if you're doing induction.
If you (for some reason) want to avoid tori $\mathbb F_q^{\times}$ and twisted products of non-trivial groups in this definition, but keep $GL_n(\mathbb F_q)$, you can maybe add the assumption to $\mathcal L_3$ that the root datum associated to $\mathbf G$ is indecomposable, and the associated root system not a product of two non-trivial root systems, or something like that? To an outsider this seems artificial?
However, not all classical groups will be in $\mathcal L_3$, e.g., $SO_{2n}^+(\mathbb F_q)$ (with the standard (classical!) definition). The problem arises from that its not connected as an algebraic group when $q$ is a power of two, so one has to pass to an index 2 subgroup, the kernel of the "pseudo/quasi-determinant", instead of the old-fashioned determinant. In particular the Weyl group of $E_8$, which is a central extension $2\cdot SO_8^+(8)$, is also not in $\mathcal L_3$, as Derek alluded to. ($W_{E_8}$ is "even further" freakish than $SO_8^+(2)$ since the central extension is related to that the index two subgroup $Spin^+_8(2)$ has non-generic Schur multiplier $\mathbb Z/2 \times \mathbb Z/2$, not zero as usual.)
I would (perhaps naively) argue that these groups also should not be considered finite groups of Lie type since they (unless I'm mistaken) do not admit the structure of a BN-pair; the structure of their p-radical subgroups is more chaotic (the Borel-Tits theorem fails).
Interpreting your $\mathcal L_2$ to consist of groups $G$ which has a filtration by normal subgroups $1< Z < H < G$, where Z is central in $G$, $G/H$ is abelian, and $H/Z$ is a simple non-abelian subquotient of a group in $\mathcal L_3$ (or equivalently $\mathcal L_1$), we will have $\mathcal L_3 \subseteq \mathcal L_2$.
With these interpretations, we can give the following response to your question:
$\mathcal L_1 \subseteq \mathcal L_3 \subseteq \mathcal L_2$, with $W_{E_8}$ or $SO_{2n}^+(\mathbb F_q)$, $q$ even, being examples of groups in $\mathcal L_2$ not in $\mathcal L_3$.
Note that a group like $GL_n(\mathbb F_p)$ of course would be in $\mathcal L_3$. Whether it is in $\mathcal L_1$ depends a bit on which $\mathbf G$ you are willing to start with.
Now, you may want to limit the class in $\mathcal L_2$ in some way to ensure that it is not larger than $\mathcal L_3$ and avoid the "pathologies". I would be surprised if there was a good way to do this, without referring back to $\mathcal L_3$: The main problem, I think, it that there is no really good "purely group theoretic" way of e.g., specifying that you only allow extensions "on top" by diagonal automorphisms (not field or graph), without getting back into the Lie theory. This is at least what goes wrong in the $SO_{2n}^+(\mathbb F_q)$, $q$ even, case. (Derek also hinted at this.)
You may also, conversely, argue that $\mathcal L_2$ is anyway too small, since it is not closed under iteratively putting things "on top" and "on the bottom"; to be "almost" of Lie type should be a bigger class... The cleanest fix for this is to go all the way and study the groups which are virtually a finite group of Lie type, i.e., has a finite index subgroup which a finite group of Lie type. Now, this class of groups is however of course just the class of virtually trivial groups, aka all finite groups...
This is not as silly as it sounds, since many deep open problems in finite group theory, such as Alperin's conjecture and its cousins, stem from the philosophy that all finite groups behave almost like finite groups of Lie type....
It is also possible to use the (exceptional) outer automorphism of order $2$ of $S_{6}$ to give an "explanation" of why $S_{6}$ is not $\{2,3\}$-generated, along the lines I used in comments for $S_{5}$ above. Take a $6$-cycle $\sigma \in S_{6}.$ Then $\sigma^{2}$ is a product of two disjoint three cycles and $\sigma^{3}$ is a product of three disjoint $2$-cycles. These clearly commute. Now take an outer automorphism $\tau$ of $S_{6}$ which sends products of two disjoint three cycles to three cycles. Then $\tau$ must also send products of three disjoint transpositions to transpositions, since $\tau(\sigma^{2})$ and $\tau(\sigma^{3})$ must commute.
Now suppose that $S_{6}$ is $\{2,3\}$-generated say $S_{6} = \langle \alpha, \beta : \alpha^{2} = \beta^{3} = 1 \rangle.$ Then we may apply $\tau$ if necessary, and assume that $\beta$ is a $3$-cycle. Then $\alpha$ is an odd permutation, so is either a transposition, or a product of three disjoint transpositions.
In the former case, we have a contradiction since there is a point fixed by both $\alpha$ and $\beta$. In the latter case, none of the transpositions in $\alpha$ can fix all points moved by $\beta$, for otherwise that transposition would be central in $\langle \alpha, \beta \rangle.$ It follows that $\alpha$ sends each point moved by $\beta$ to a point fixed by $\beta$ and conversely. It follows that $\beta$ and $\beta^{\alpha}$ commute. Now $\alpha$ normalizes the Abelian subgroup $\langle \beta, \beta^{\alpha}\rangle $, so that $\langle \alpha, \beta \rangle = \langle \alpha \rangle \langle \beta^{\alpha} ,\beta \rangle$ has order dividing $18$, a contradiction.
I do not know if there is an argument using the fact that $S_{8}$ is isomorphic to ${\rm GL}(4,2)\langle \gamma \rangle$, where $\gamma$ is the transpose inverse automorphism, to "explain" that $S_{8}$ is not $\{2,3\}$-generated.
Later edit: It would have been better perhaps to use the outer automorphism of $S_{6}$ to reduce to the case that $\alpha$ is a transposition,(in which case, generation requires that $\beta$ is a product of two disjoint $3$-cycles), and then note the general fact that when $n >1$, $S_{2n}$ is never generated by a transposition $\alpha$ and an element $\beta$ which
is a product of two disjoint $n$-cycles. For if it were, we may conjugate the pair and assume that $\alpha = (12).$ If either of the $n$-cycles in $\beta$ were disjoint from $\alpha$, then that $n$-cycle would be central in $\langle \alpha, \beta \rangle = S_{2n}$, a contradiction. Hence both $n$-cycles of $\beta$ contain a point moved by $\alpha$.
We may conjugate $\beta$ by a permutation fixing both $1$ and $2$ and assume that $\beta = (1357 \ldots 2n-1)(2468 \ldots 2n)$ without disturbing the generation property. Then $\langle \alpha, \alpha^{\beta}, \ldots, \alpha^{\beta^{n-1}}\rangle$ = $\langle (12),(34), \ldots , (2n-1 2n) \rangle$ is Abelian and normal in $\langle \alpha, \beta \rangle = S_{2n},$ a contradiction.
Best Answer
V. Dotsenko's construction, on math.stackexchange:
https://math.stackexchange.com/questions/1401/why-psl-3-mathbb-f-2-cong-psl-2-mathbb-f-7/1450#1450
may fit your requirement "combinatorial mapping of these geometries that induces an isomorphism".