You're right on a lot of counts. The wavefunction of the system is indeed a function of the form
$$
\Psi=\Psi(\mathbf r_1,\mathbf r_2),
$$
and there's no separating the two, because of the cross term in the Schrödinger equation. This means that it is fundamentally impossible to ask for things like "the probability amplitude for electron 1", because that depends on the position of electron 2. So at least a priori you're in a huge pickle.
The way we solve this is, to a large extent, to try to pretend that this isn't an issue - and somewhat surprisingly, it tends to work! For example, it would be really nice if the electronic dynamics were just completely decoupled from each other:
$$
\Psi(\mathbf r_1,\mathbf r_2)=\psi_1(\mathbf r_1)\psi_2(\mathbf r_2),
$$
so you could have legitimate (independent) probability amplitudes for the position of each of the electrons, and so on. In practice this is not quite possible because the electron indistinguishability requires you to use an antisymmetric wavefunction:
$$
\Psi(\mathbf r_1,\mathbf r_2)=\frac{\psi_1(\mathbf r_1)\psi_2(\mathbf r_2)-\psi_2(\mathbf r_1)\psi_1(\mathbf r_2)}{\sqrt{2}}.
\tag1
$$
Suppose that the eigenfunction was actually of this form. What could you do to obtain this eigenstate? As a fist go, you can solve the independent hydrogenic problems and pretend that you're done, but you're missing the electron-electron repulsion. You could solve the hydrogenic problem for electron 1 and then put in its charge density for electron 2 and solve its single electron Schrödinger equation, but then you'd need to go back to electron 1 with your $\psi_2$. You can then try and repeat this procedure for a long time and see if you get something sensible.
Alternatively, you could try reasonable guesses for $\psi_1$ and $\psi_2$ with some variable parameters, and then try and find the minimum of $⟨\Psi|H|\Psi⟩$ over those parameters, in the hope that this minimum will get you relatively close to the ground state.
These, and similar, are the core of the Hartree-Fock methods. They make the fundamental assumption that the electronic wavefunction is as separable as it can be - a single Slater determinant, as in equation $(1)$ - and try to make that work as well as possible. Somewhat surprisingly, perhaps, this can be really quite close for many intents and purposes. (In other situations, of course, it can fail catastrophically!)
In reality, of course, there's a lot more to take into account. For one, Hartree-Fock approximations generally don't account for 'electron correlation' which is a fuzzy term but essentially refers to terms of the form $⟨\psi_1\otimes\psi_2| r_{12}^{-1} |\psi_2\otimes\psi_1⟩$. More importantly, there is no guarantee that the system will be in a single configuration (i.e. a single Slater determinant), and in general your eigenstate could be a nontrivial superposition of many different configurations. This is a particular worry in molecules, but it's also required for a quantitatively correct description of atoms.
If you want to go down that route, it's called quantum chemistry, and it is a huge field. In general, the name of the game is to find a basis of one-electron orbitals which will be nice to work with, and then get to work intensively by numerically diagonalizing the many-electron hamiltonian in that basis, with a multitude of methods to deal with multi-configuration effects. As the size of the basis increases (and potentially as you increase the 'amount of correlation' you include), the eigenstates / eigenenergies should converge to the true values.
Having said that, configurations like $(1)$ are still very useful ingredients of quantitative descriptions, and in general each eigenstate will be dominated by a single configuration. This is the sort of thing we mean when we say things like
the lithium ground state has two electrons in the 1s shell and one in the 2s shell
which more practically says that there exist wavefunctions $\psi_{1s}$ and $\psi_{2s}$ such that (once you account for spin) the corresponding Slater determinant is a good approximation to the true eigenstate. This is what makes the shells and the hydrogenic-style orbitals useful in a many-electron setting.
However, a word to the wise: orbitals are completely fictional concepts. That is, they are unphysical and they are completely inaccessible to any possible measurement. (Instead, it is only the full $N$-electron wavefunction that is available to experiment.)
To see this, consider the state $(1)$ and transform it by substituting the wavefunctions $\psi_j$ by $\psi_1\pm\psi_2$:
\begin{align}
\Psi'(\mathbf r_1,\mathbf r_2)
&=\frac{\psi_1'(\mathbf r_1)\psi_2'(\mathbf r_2)-\psi_2'(\mathbf r_1)\psi_1'(\mathbf r_2)}{\sqrt{2}}
\\&=\frac{
(\psi_1(\mathbf r_1)-\psi_2(\mathbf r_1))(\psi_1(\mathbf r_2)+\psi_2(\mathbf r_2))
-(\psi_1(\mathbf r_1)+\psi_2(\mathbf r_1))(\psi_1(\mathbf r_2)-\psi_2(\mathbf r_2))
}{2\sqrt{2}}
\\&=\frac{\psi_1(\mathbf r_1)\psi_2(\mathbf r_2)-\psi_2(\mathbf r_1)\psi_1(\mathbf r_2)}{\sqrt{2}}
\\&=\Psi(\mathbf r_1,\mathbf r_2).
\end{align}
That is, the Slater determinant that comes from linear combinations of the $\psi_j$ is indistinguishable from the one you get from the $\psi_j$ themselves. This extends to any basis change on that subspace with unit determinant; for more details see this thread. The implication is that labels like s, p, d, f, and so on are useful to describe the basis functions that we use to build the dominating configuration in a state, but they cannot be reliably inferred from the many-electron wavefunction itself. (This is as opposed to term symbols, which describe the global angular momentum characteristics of the eigenstate, and which can indeed be obtained from the many-electron eigenfunction.)
Best Answer
The energy levels depend on two things:
the electrostatic attraction between the electrons and the nucleus
the electrostatic repulsion between the electrons
If you take a hydrogen atom, which is what your diagram shows, then there is a single electron and a single proton. The electron is attracted to the proton and there is no electron-electron repulsion because there is only one electron.
If you move on the the next element, helium, there are two electrons and the nucleus contains two protons. So the attraction between the electrons and the nucleus is now twice as big but we have a repulsion between the two electrons. Both these factors change the energy levels so they are not the same as hydrogen. The next element, Lithium, has three electrons and three protons in the nucleus so the energy levels are different again. And so on.
So all the atoms of a given element have the same energy levels because they have the same numbers of electrons and protons. For example all hydrogen atoms have the same energy levels. But the different elements have different energy levels because they contain different numbers of electrons and protons. The hydrogen energy levels differ from helium, which in turn differs from lithium and so on.
And just to complicate matters the number of neutrons in the nucleus makes a small difference as well, so for example the energy levels of hydrogen are slightly different from the energy levels of deuterium and tritium.