Well, here is a seat-of-the pants "lark" formal answer, going to "rigged" Fock spaces and places you (or anybody else) shouldn't really be at; except you may already have been there, when you learned about bras and kets, if you looked in Dirac's QM book ― mysteriously, the section virtually all modern texts skip!
I will cover your Fock space question, but mostly as an introduction to segue into Dirac's sublime picture that merits exposure. In this sense, this informal formal answer is exceptionally indulgent.
Indeed, for coherent states,
$$
|\alpha\rangle= e^{-|\alpha|^2/2 } e^{\alpha a^\dagger}|0\rangle ,
$$
where $a|0\rangle =0$ and $[a,a^\dagger]=1$, you find that $a|\alpha\rangle=\alpha | \alpha\rangle$.
The adjoint/dual relation in your (1) is nothing but these, so they really do not tell you something new, and they are not the "left" eigenstates of $a^\dagger$ in any meaningful sense, except the trivial one. You are just looking at the very same states in dual space.
The strict proofs of linked questions excluding eigenstates of $a^\dagger$ have a flakey formal loophole, however. The freak (unnormalizable) state
$$
|\psi\rangle=\delta(a^\dagger - \beta ~ \mathbb{I}) |0\rangle =\frac{1}{2\pi} \int \!\! dk ~e^{ik(a^\dagger -\beta)} |0\rangle \\
=\frac{1}{2\pi}\int\!\! dk~\exp(-ik\beta) \left (|0\rangle+ ik |1\rangle + ... +(ik)^n/\sqrt{n!}|n\rangle +...\right ) \\
= \delta(\beta)|0\rangle -\partial_\beta \delta (\beta)|1\rangle +...+{(-\partial_\beta )^n \delta(\beta)\over \sqrt{n!}}|n\rangle+...
$$
formally satisfies
$$
a^\dagger | \psi\rangle=\beta| \psi\rangle,
$$
and is the sought-after right-eigenstate of $a^\dagger$, so, if you create/add an excitation, it indeed does not change, as per your (2).
It is essentially a formal continuous superposition of an infinity of tilted coherent states, analogous to Dirac's $|x\rangle$, which is where it came from, of course; see below.
I do not know if these states have actually entered in some twisted technical fashion into quantum optics as a tool, but, frankly, I am terminating this part of the discussion, regarding it as the warmup handle to the beautiful theory of Dirac's kets, which he details in his classic QM book (4th ed.), Ch III §20 & Ch IV §22, §23.
Most of the formal maneuvers we've seen so far algebraically map in a popular loose analogy into the standard Dirac bra-ket entities and operators,
$$
a^\dagger \rightsquigarrow \hat x , \qquad a \rightsquigarrow \hat p , \qquad [a^\dagger,a]=1 \rightsquigarrow [\hat x , \hat p]= i\hbar.
$$
Let's forget about creators and annihilators, if they confuse you, and start with Dirac's standard ket,
$$
\rangle\equiv \sqrt{2\pi\hbar } |\varpi\rangle.
$$
I introduce my own notation on the r.h.s. to mitigate the culture shock which has alienated generations: What I mean by it is the standard translationally invariant vacuum, an infinite x-vector with the same constant component in every entry, so that a translation leaves it invariant. (Since you know the tail end of this notation already, it will turn out to be $\lim_{p\to 0} |p\rangle\equiv |\varpi\rangle$. I will not use 0 instead of $\varpi$, as I want to be reminding you it is a p-eigenstate and not an x-eigenstate. So, in this picture, $\hat p$s are annihilators and $\hat x$s creators.)
It is thus the analog of the coherent state vacuum, the null state of
of $\hat p$,
$$ \bbox[yellow,5px]{
\hat p |\varpi\rangle = 0}.
$$
Dirac then defines
$$ \bbox[yellow,5px]{
|x\rangle = \delta(\hat x -x)|\varpi \rangle \sqrt{2\pi \hbar}},
$$
and
$$ \bbox[yellow,5px]{
|p\rangle = e^{ip\hat x /\hbar} | \varpi \rangle},
$$
a momentum-translated zero momentum state, $e^{p \partial_{\varpi}} | \varpi \rangle$, the analog of the coherent state!
Check that, therefore,
$$
\langle x | p\rangle = \sqrt{2\pi \hbar} \langle \varpi | \delta (\hat x - x ) e^{ip\hat x /\hbar} |\varpi\rangle =e^{ip x /\hbar} \langle x |\varpi\rangle = e^{ip x /\hbar} /\sqrt{2\pi \hbar},
$$
since the projection of all x-eigenstates on the standard ket is the same, and
$$
\hat p |p\rangle = \hat p e^{ip\hat x /\hbar}|\varpi\rangle = p|p\rangle , \qquad \hat x |x\rangle = \hat x \delta(\hat x -x)|\varpi \rangle \sqrt{2\pi \hbar}=x |x\rangle,
$$
the properties that most QM textbooks start with.
(But, to my surprise when I first learned this, Dirac evidently understood the structure of coherent states a long time before their official inception...)
If you must have a mental picture of these infinite-dimensional vectors in the x-basis, the $|x\rangle$ vector only has a nonzero entry in the x=n th slot, so is terminally sharp; but the $|\varpi \rangle$ vector is terminally broad, and has, say, 1 in every single entry, suitably normalized by the continuum normalization of these monsters; and the $|p\rangle $ vector is as broad, and has entries around the 0th slot which go like $...,e^{-2ip}, e^{-ip}, 1,e^{ip},e^{2ip},e^{3ip},...$. Afficionados of Finite Fourier Transform will recognize these as akin to beloved vectors.
NB Add-on addressing the no-go proof
Through the orgy of distributions involved, one may see the strictly formal loophole of the no-go proof,
$$
a^\dagger \left (\delta(\beta)|0\rangle -\partial_\beta \delta (\beta)|1\rangle +...+{(-\partial_\beta )^n \over \sqrt{n!}}\delta(\beta)|n\rangle+... \right ) \\ =
\beta \left ( \delta(\beta)|0\rangle -\partial_\beta \delta (\beta)|1\rangle +...+{(-\partial_\beta )^n \over \sqrt{n!}}\delta(\beta)|n\rangle+...\right ).
$$
Note the "snag" $|0\rangle$ term on the right vanishes by virtue of $\beta \delta(\beta)=0$, the second term $-\beta \partial_\beta \delta(\beta) = \delta (\beta)$, etc. Hyperformal to be sure, but operator-valued distributions are the lifeblood of QFT, and the task of "cleaning up" the landscape undertaken by axiomatic QFT is still not complete... In QM people have come to expect some rigging stunts will settle everything, but I am not too good at that.
The discovery and study of coherent states represents one aspect of one
of the biggest problems physicists have faced with the birth and the
subsequent development, supported by excellent experimental results, of the
quantum mechanics: the search for a correspondence between the new theory, conceived
for the analysis of microscopic systems, and classical physics, still fully valid
for the description of the macroscopic world.
The history of coherent states begins immediately after the advent of mechanics
quantum: their introduction on a conceptual level dates back to a
article published in 1926, in which Schrödinger reports the existence of a class of
states of the harmonic oscillator that show, in a certain sense, behavior
analogous to that of a classic oscillator: for these states it is verified that the energy
mean corresponds to the classical value and the position and momentum averages have
oscillatory forms in constant phase relation.
Returning to Schrödinger's article, the "almost classical" states from him
identified present, in addition to the characteristics already mentioned, an important aspect:
being represented by Gaussian wave packets that do not change shape in the
time, guarantee the minimization of the product among the uncertainties about
position and on the impulse, that is the condition closest to the possibility of measuring
simultaneously the aforesaid quantities with arbitrary precision, allowed
from classical physics.
So, starting from the following relations:
\begin{equation}
a^{\dagger}|n\rangle=\sqrt{n+1}|n+1\rangle \quad a|n\rangle=\sqrt{n}|n-1\rangle
\end{equation}
it is noted that, by virtue of the orthonormality of the states
stationary, the diagonal matrix elements of the position and momentum operators are
null in the representation of energy, which means that the expectation values of
position and momentum on any stationary state are zero instant by instant.
The stationary states just analyzed are characterized by distributions of
constant probabilities with respect to the position over time; the wait values of the position e
of the impulse are null at all times: this aspect is a fundamental one
difference with the states of the classic oscillator, for which, once the energy is defined
(as long as different from zero), the observables position and momentum evolve over time
according to sinusoidal functions and are always in phase quadrature with each other. Also, if yes
calculate the uncertainties on position and momentum for a steady state with n photons, yes
gets the uncertainty relation $\Delta x \Delta p=(n+1/2)\hbar$.
it is therefore possible to obtain the minimization of the product of the uncertainties on impulse and
position, which represents the maximum similarity with classical mechanics.
A state that is as similar as possible to the classical case must therefore
have the following characteristics:
1): The evolution over time of the position and momentum expectation values must
be of a simple periodic type, with a constant phase ratio between position and
impulse.
2): The wave functions must be as narrow as possible around the value
average of the position, so that the probability distribution with respect to the
position may tend, by varying appropriate parameters, to a delta function of
Dirac;
3): The product of the uncertainties on the position and on the impulse must be minimal.
So you can see this "classical" behavior, that admits these particular states, as an intrinsic property of QHO.
Furthermore I have say that: also the harmonic oscillator energy eigenstates actually behave like oscillators.
Maybe this answer is a bit too long but I hope it can help you.
Best Answer
Derivation from the eigenvalue condition
The most straightforward approach is to start from the position representation of the annihilation operator $a$.
I'll use the convention $a=\frac{1}{\sqrt2}(x+ip)$ and $a^\dagger=\frac{1}{\sqrt2}(x-ip)$, corresponding to which the canonical commutation relations read $[a,a^\dagger]=1$ and $[x,p]=i$. In the position representation, this corresponds to $a=\frac{1}{\sqrt2}(x+\partial_x)$.
The eigenvalue condition $a|\alpha\rangle=\alpha|\alpha\rangle$ then corresponds to $$x\psi_\alpha(x)+\psi_\alpha'(x)=\sqrt2\alpha\psi_\alpha(x),$$ which we can rewrite as $$\psi_\alpha'(x) = (\sqrt2\alpha-x)\psi_\alpha(x).$$ A natural ansatz to solve this is $\psi_\alpha(x)=C e^{f(x)}$ for some constant $C$. Using this we get $$f'(x) = \sqrt2\alpha-x \Longrightarrow f(x) = \sqrt2\alpha x - \frac{x^2}{2} + C'.$$ Upon some simple rearranging of the terms we get $\psi_\alpha(x) \propto \exp[-\frac12(x - \sqrt2\alpha)^2]$, which ensuring normalisation finally leads to $$\psi_\alpha(x) = \frac{e^{-\alpha_2^2}}{\pi^{1/4}} \exp\left[-\frac12(x - \sqrt2\alpha)^2\right],\tag{$R_1$}$$ where $\alpha=\alpha_1+i\alpha_2$.
Derivation from (A)
Consider the wavefunctions $\psi_n$ of $|n\rangle$, which have the form $$\psi_n(x) = \frac{1}{\pi^{1/4}\sqrt{2^n n!}} e^{-x^2/2}H_n(x),\tag B$$ where $H_n$ are the Hermite polynomials, defined here as $H_n(x)=(2x-\partial_x)^n \cdot 1$. This expression for $\psi_n$ comes from $$\psi_n(x) = \langle x|\frac{a^{\dagger n}}{\sqrt{n!}}|0\rangle = \frac{1}{\pi^{1/4}\sqrt{2^n n!}} (x-\partial_x)^n e^{-x^2/2} = \frac{1}{\pi^{1/4}\sqrt{2^n n!}} e^{-x^2/2}H_n(x).$$ Using (B) in (A), we get $$ \psi_\alpha(x) = e^{-|\alpha|^2/2} \frac{e^{-x^2/2}}{\pi^{1/4}} \sum_{k=0}^\infty \frac{\alpha^k}{\sqrt{k!}} \frac{1}{\sqrt{2^k k!}} H_k(x). $$ We now consider the identity $$\sum_{k=0}^\infty H_k(x) \frac{t^k}{k!} = e^{2xt - t^2}.$$ Using this with $t=\alpha/\sqrt2$ we get $$\psi_\alpha(x) = e^{-|\alpha|^2/2} \frac{e^{-x^2/2}}{\pi^{1/4}} e^{\sqrt2\alpha x - \alpha^2/2} = \frac{1}{\pi^{1/4}}e^{\frac12(\alpha^2-|\alpha|^2)}\exp\left[ -\frac12(x-\sqrt2\alpha)^2 \right]. \tag{$R_2$} $$ Observing that $\alpha^2-|\alpha|^2=-2\alpha_2^2 + 2i\alpha_1\alpha_2$, we see that ($R_2$) is consistent with ($R_1$), up to an irrelevant global phase. An analogous derivation is also given in Gerry, Knight (2004), section 3.3.