There's an infinite number of orbitals, and thus an infinite number of possible state transitions. However, the energy of the orbitals asymptotes to a finite value - for example, a hydrogen atom has energy levels given by the formula $E_n = -\frac{13.6\text{ eV}}{n^2}$ (ignoring some very tiny quantum corrections). If you let $n$ go up to infinity, the energy approaches zero. So a photon with enough energy, namely 13.6 eV, will knock the electron past all those infinite energy levels and out of the atom entirely.
Furthermore, if you look at some of the numerical values that come out of that formula: -13.6 eV, -3.4 eV, -1.51 eV, -0.85 eV, -0.54 eV, etc., you'll notice that they get closer and closer together as $n$ gets higher. At some point, the difference between one energy level and the next becomes so small that you can't measure it.
All instances of a particular atom are exactly the same, and so a given transition will always have the same energy for a particular kind of atom (including the isotope). This is how we are able to identify elements by their spectral lines. However, different atoms have different energy levels, because of the different nuclear charge and mass, and also because of multiple electron interactions, so the energy will vary from one atom to another, even for the same transition.
To address the question in your title, there is a list of the spectral lines of most elements in the CRC Handbook of Chemistry and Physics. I'm not sure if it lists the corresponding state transitions, though. Unfortunately the CRC website itself is restricted, so you'd need to pay to access it, but the table itself is probably available elsewhere online. It's quite extensive, listing hundreds of spectral lines for each of the elements.
Based on some of your comments, I think what might be tripping you up is the first statement you started with:
From the Bohr's atomic model, it is clear that electron can have only certain definite energy levels.
and
...If suppose, we assume electron losses total energy, electron can't stay in any particular shell, as it would not have that particular value of energy.
That may be true for Bohr's atomic model, but Bohr's atomic model is wrong. And electron does not have to be in a particular, definite shell or energy level. Rather, any electron state is a superposition of states of definite energy level (energy eigenstates).
That means the expectation value of a hydrogen electron state is going to look like
$$\langle E\rangle = \sum_n |a_n|^2 E_n\text{,}$$
where $\{a_n\}$ are arbitrary complex values with $\sum_{n>0}|a_n|^2 = 1$ and $E_n$ are energy levels in increasing order. Because of the sum-to-$1$ condition, taking any portion along the other energy eigenstates will increase energy compared to the ground state.
In other words, even if the electron state does not have a definite energy, you still can't go lower than the ground state.
Suppose, I have a cup of hot coffee on the table. It will be continuously losing energy in the form of heat, but it stays on the table, though there was a energy loss. Now, all of a sudden, I take off the table, the cup of coffee converts it potential energy into kinetic energy to come down.
If you don't shake the table, the coffee cup will sit there, forever. Similarly, nothing perturbs the electron in an excited energy eigenstate, then it simply will never decay. It cannot: energy eigenstates are stationary; they do not evolve into anything other than themselves.
However, being completely without external perturbation is actually impossible. The uncertainty principle provides the electromagnetic field with vacuum fluctuations, which will perturb the electron even if nothing else in the environment does. In your analogy, this (or something else) provides the "shaking of the table" for the electron. Once the electron state gains even a tiny component in some other energy eigenstate, the state can evolve in time.
In other words, one can think of spontaneous emission as a particular type of stimulated emission where it's the vacuum that does the perturbing.
Best Answer
Multielectron atom has much more complex energy spectrum than hydrogen atom. As the electrons interact with each other, the hydrogenic energy levels get shifted, and much of the hydrogen-specific degeneracy, as well as degeneracy resulting from electrons mass&charge equality, is lifted. Moreover, since the electrons do interact with each other, we can't, strictly speaking, speak about single-particle states of electrons. This means that when we shine light on a multielectron atom, it can't excite one electron. It instead changes the configuration of the whole atom, and the final state is approximately what we would call an atom with single electron excited.
Now, as the energy levels are partially split by electron-electron interaction, there's a way$^\dagger$ to directly (i.e. via single photon) get to the state, which we approximately (i.e. in terms of the shell model) describe as twice-excited state. We just have to shine the appropriate frequency of light at the atom. This way we could even excite more than two electrons. Doing another way, we could shine light composed of two frequencies corresponding to transition from ground to singly excited state and from singly excited to doubly excited state. Then it'll also be possible for the atom to appear in doubly excited state, but this would be a sequential transition.
For some atoms though, doubly excited states are unstable in the sense that the corresponding energy is above ionization energy, so such states are likely to decay into ionized atom instead of directly going to ground state. This is true for e.g. doubly excited helium atom.
Note that transition from one state to another is not instant. The probability of occupying some state gradually changes with time, and if you shine the light driving transition to singly-excited state and mix the light driving transition from singly excited to doubly excited state into the same beam, you'll be able to "intercept" the atom's singly excited state and drive it to doubly excited one.
Don't forget about quantization of energy and the fact that electron-electron interaction lifts much of the possible degeneracy. It's highly unlikely that energies of doubly excited atom and atom with single electron excited to higher level will coincide. Thus if you tune your light source well enough, you're unlikely to excite the already excited electron to higher level instead of getting doubly excited atom.
$^\dagger$ I may actually be wrong if it appears that such optical transition is forbidden for all atoms. If you find some evidence of this, please let me know.