If Alice and Bob witness two light flashes $E1$ and $E2$ [...]

Apparently were considering two inertial frames, with Alice being a member of one of these two, and Bob being a member of the other;

and two events (which are in turn observable, or "flashing") such that one (unnamed) member of Alice's frame and one (unnamed) member of Bob's frame were participating (in coincidence, passing each other) in one of these event, and another (though, in some cases, not necessarily distinct from the aforementioned) (unnamed) member of Alice's frame and another (though, in some cases, not necessarily distinct from the aforementioned) (unnamed) member of Bob's frame were participating (in coincidence, passing each other) in the other event; and further,

• that the distance between these two identified members of Alice's inertial frame is denoted as $\Delta x_A$ (and that these two members are also called "the two ends of Alice's measuring stick"),

• that the distance between these two identified members of Bob's inertial frame is denoted as $\Delta x_B$ (and that these two members are also called "the two ends of Bob's measuring stick"),

• that the duration of one of the end's of Alice's "measuring stick" from its instant of having taken part in one of the two mentioned events (having met and passed one of the ends of Bob's "measuring stick"), until its instant simultaneous to the instant of the other end of Alice's "measuring stick" having taken part in the other event (having met and passed the other end of Bob's "measuring stick") is denoted as $\Delta t_A$, and

• that the duration of one of the end's of Bob's "measuring stick" from its instant of having taken part in one of the two mentioned events (having met and passed one of the ends of Alice's "measuring stick"), until its instant simultaneous to the instant of the other end of Bob's "measuring stick" having taken part in the other event (having met and passed the other end of Alice's "measuring stick") is denoted as $\Delta t_B$,

then $(\Delta x_B)^2 - c^2(\Delta t_B)^2 = (\Delta x_A)^2 - c^2(\Delta t_A)^2$

Right; where $c$ symbolizes signal front speed, of course. (My slightly verbose explanation of all symbols in your equation was to emphasize that, despite appearance, it is a coordinate-free statement.)

Let's assume that Alice observes these flashes at each end of a measuring stick, which is at rest with respect to Alice, both at the same time in her reference frame.

Let's (correspondigly) call the two "flash" instants or indications of the two ends of Alice's "measuring stick" simultaneous to each other (according to Einstein's definition of how this ougth to be measured).

Then $\Delta t_A = 0$, and we get $(\Delta x_B)^2 - c^2(\Delta t_B)^2 = (\Delta x_A)^2$.

Correct.

We assume now that Alice (and, hence, the measuring stick) are moving at a non-0 velocity $vB$ when observed by Bob.

Bob and all relevant members of Bob's inertial frame measure

• that Alice and all members of Alice's inertial frame move uniformly wrt. Bob's inertial frame, and

• that they all move at equal (and, necessarily constant) non-zero speed $vB$.

(Also, Alice and the members of Alice's frame can determine vice versa the constant speed $vA$ of Bob and all members of Bob's inertial frame; and they find out that $vA = vB$.)

_Now [...] factoring out $(\Delta x_B)^2$

... yes ...

[we get] $(\Delta x_B)^2(1 - \frac{c^2}{v_B^2}) = (\Delta x_A)^2$

No! -- $vB$ is certainly not defined as ratio between $(\Delta x_B)$ and $(\Delta t_B)$!

(And also, generally the values of these two differently defined quantities are not equal.)

Instead, with $\Delta t_A = 0$, it can be derived (by other, more elementary arguments) that

$$(\Delta x_B)^2 - c^2(\Delta t_B)^2 = (\Delta x_A)^2 = (\Delta x_B)^2 \, \left( 1 - \frac{vB^2}{c^2} \right),$$

a.k.a. "length contraction".