Generally, the $1s$ and $2s$ orbitals will be associated with different energies in different atoms, which means that a transition that moves an electron from the $2s$ orbital to the $1s$ orbital will emit at a different wavelength for different atoms. It is unclear what you mean with "type of photon", but having different photon energy definitely qualifies. (On the other hand, the angular-momentum characteristics of those photons ─ specifically, their polarization and spatial profile ─ would be the same.)
However, some important warnings:
The $2s\to 1s$ transition is dipole-forbidden, which means that it will not be present unless you make a strong and dedicated effort to suss it out.
(Though that said, the $2s\to1s$ transition is an important staple of the precision spectroscopy of the hydrogen atom (example), since its dipole-forbidden nature means that its natural lifetime is much longer than dipole-allowed lines, which therefore means that its natural linewidth is correspondingly narrower.)
Saying "$2s\to 1s$ transition" intrinsically assumes that you have a one-electron atom, and that there are no other electrons to keep track of. This is obviously only true in hydrogen and hydrogenic ions like $\rm He^+$, $\rm Li^{2+}$, $\rm Be^{3+}$, and a series of increasingly unlikely and unavailable highly-charged species.
In general, however, you can't just say that - you need to fully specify the initial and final electronic configurations: thus, you'd say $1s^12s^1\to1s^2$ in helium, for example, or $1s^12s^2\to1s^22s^1$ in lithium, say.
If you go beyond that, though, you run into the problem that you cannot move an electron from the $2s$ shell to the $1s$ shell without this being a transition between two excited states: starting from beryllium, the only way to have such a transition would be to do something like
$$1s^1 2s^2 2p^1 \to 1s^22s^1 2p^1, \tag{$*$}$$
and that's already impossible: the second state $1s^22s^1 2p^1$, does exist as the first excited state of beryllium, but the core hole implied by the $1s^1$ in $1s^1 2s^2 2p^1$ requires more energy to create than it takes to just remove one of the $2s^2$ electrons, which means that the initial state is an unstable autoionizing state with a very short lifetime, and it doesn't exist in nature. (You can go look for the energy levels of beryllium in the NIST ASD - ask for Be I in the search box - and confirm that there is no such state reported there.) The transition in $(*)$ might be accessible as a resonance in attosecond transient absorption experiments (here/here is a similar example in helium), but that's about it.
To address your other questions:
Also, does a change from a higher principal quantum number to a lower one always produce a high-energy photon? Always higher than a drop in sublevels ($p$ to $s$, e.g.)?
Generally speaking, yes: transitions which change the principal quantum number of a valence electron generally have higher transition energies than transitions between subshells of different angular momentum which keep the principal quantum number intact. That's not a universal fact, though - as atoms get large, there's all sorts of wonky phenomena that can kick in.
In other words, is the amount of energy (in electron-volts) between $2s$ and $2p$ the same or similar for different chemical species?
This is a pretty universal rule: you'll basically never find two transitions in two different chemical species which have identical transition energies; this is what allows spectroscopy to work, since it allows emission and absorption lines to function as chemical fingerprints.
For $2p\to 2s$ transitions, it's probably best if you go and have a play yourself using the NIST ASD, on either the lines or the levels databases for the first-period $p$-block elements (boron through fluorine; add I to get the neutral atom).
If an electron changes spin direction (the fourth quantum number) does that produce a weak photon?
Transitions which change spin direction are electric-dipole-forbidden, and they can only take place from magnetic-dipole onwards, which means that they're suppressed (in likelihood, not in energy) compared to transitions which don't flip any spins. If the only thing you're doing is to flip a spin, then there are two possibilities:
You're flipping the spin of an electron in isolation, such as e.g. you have a helium ion $\rm He^+$ in the $1s$ state, and you're changing the direction of the spin of the lone electron. In this type of situation, the atomic hamiltonian is isotropic, so it doesn't care about the direction of the spin, and if your atom is unperturbed then the transition energy will be exactly zero. When we do this, however, we are generally applying an external magnetic field, which will interact with the electron spin via the Zeeman effect, and that will induce a small energy difference between the two levels, typically in the radio-frequency regime.
It's also possible that you're flipping the spin of an electron with respect to some other system: either the nucleus (as in atomic hydrogen, say) or other electrons. These represent transitions within the hyperfine-structure and fine-structure manifolds of each level, which tend to be low-energy transitions, ranging from the radio regime to the high microwave bands depending on exactly what you're dealing with.
Generally, saying "weak photon" is unclear language and it should be avoided. In both of the cases above, the photons will have a low photon energy, and they will also be dipole-forbidden, which means that they will be relatively unlikely. (However, there's plenty of important examples in this class, starting with the hydrogen 21cm line.)
If, on the other hand, you're doing a spin flip in addition to some other change (say, the $1s2p\to 1s^2$ transition in helium, where you're also flipping one of the spins, i.e. going from the spin triplet $^3P$ state down to the spin singlet ground state) then the effect will be a minute change in photon energy (in that case, of the order of 1% relative shift) and a strong suppression (here, by a factor of $10^7$) of the transition probability.
