Annihilating coherent state
What you are arguing is that you can take a coherent state $\vert\alpha\rangle$, which has energy $|\alpha|^2$ (in units of $\hbar\omega$), remove one photon which has energy $1$, and still be left with $a\vert\alpha\rangle=\vert\alpha\rangle$. Thus, you would have gained $1$ unit of energy (and you could continue doing so).
The problem is that applying $a$ is not a process you can carry out physically (i.e., deterministically on any input state). If you actually try to remove one photon from a coherent state, other things will happen.
Firstly, in order to describe the process in a clean way, you will need another system into which you transfer the energy. This could be another bosonic mode $b$ (with same energy per photon) or e.g. a two-level system. Let us first consider a bosonic mode.
One possibility to transfer the photon to $b$ would be to build a unitary doing so, such as $U=b^\dagger a$. However, you can see easily that this would not be a unitary -- for the very least, you would have to add a normalization $1/\sqrt{a^\dagger a}$, which will clearly change its effect on a coherent state.
Another approach would be to build a Hamiltonian which moves photons from $a$ to $b$, such as $H=a^\dagger b + \mathrm{h.c.}$. If you time-evolve under this Hamiltonian, what you implement is a beam splitter. It is well known that applying a beam splitter on a coherent state gives you two coherent states with the same total energy. In particular, the state $\vert\alpha\rangle$ is not preserved under the action of a beam splitter. Mathematically, this related to the fact that $\exp(iHt)$ does not only contain $a$ but all powers of $a$ and combinations with $a^\dagger$s.
Alternatively, you could try to couple your coherent state to a two-level system (described by $\sigma^+$ and $\sigma^-$ with the same energy), again using $H=\sigma^+ a + \mathrm{h.c.}$. This is a Jaynes-Cummings Hamiltonian which will induce Rabi oscillations $$ \cos(\sqrt{n+1}t)\vert n+1,g\rangle + \sin(\sqrt{n+1}t)\vert n,e\rangle $$ in each of the subspaces spanned by $\{\vert n+1,g\rangle,\vert n,e\rangle\}$ independently. (The second part is the atomic state: ground and excited.) It is clear that when you act on a coherent state (or in fact any superposition of all number states), you will never end up with a factorized state between the two subsystems, since the $\sqrt{n+1}t$ are generally incommensurate. Thus, there is no simple analysis. However, you can easily convince yourself that the energy at each instant is preserved; or you can try to find times $t$ where the system is approximately excited and check that the energy is approximately conserved for the joint energy of the two subsystems.
Can someone point me where I am wrong?
Consider a state that is a superposition of number states $|n\rangle$ for all $n$.
Recalling that
$$\hat a |n\rangle = \sqrt{n}|(n-1)\rangle$$
it's clear that
$$\hat a\left(A\sum_{n=0}^\infty \frac{|n\rangle}{\sqrt{n!}}\right) = A\sum_{n=1}^\infty \frac{\sqrt{n}|(n-1)\rangle}{\sqrt{n!}}=A\sum_{n=1}^\infty \frac{|(n-1)\rangle}{\sqrt{(n-1)!}}=A\sum_{m=0}^\infty \frac{|m\rangle}{\sqrt{m!}}$$
In a certain sense, the annihilation operation removed an infinity of quanta, one from each state in the (infinite) superposition of number states.
Since there is no highest number state, the resulting state is still a superposition of all number states and, by choosing the coefficient for each number state in the superposition to be $\frac{1}{\sqrt{n!}}$, the resulting state is then identical to the original state.