Write an arbitrary state as
$$|\Psi\rangle = \sum_{n=0}^{\infty} c_n |n\rangle \,.$$
Now apply the raising operator
$$
\begin{align}
a^\dagger |\Psi\rangle &=
a^\dagger \sum_{n=0}^{\infty} c_n |n\rangle \\
&= \sum_{n=0}^{\infty} c_n \sqrt{n+1} |n+1\rangle \\
&= \sum_{n=1}^{\infty} c_{n-1} \sqrt{n} |n\rangle
\end{align}
$$
If $|\Psi\rangle$ is an eigenstate of $a^\dagger$ with eigenvalue $\alpha$ then we have
$$\sum_{n=0}^{\infty}c_n|n\rangle = \sum_{n=1}^\infty c_{n-1} \sqrt{n}|n \rangle \, .$$
You already got this far.
Indeed, the only solution to this equation is $c_n=0$ for all $n$.
Therefore, there is no eigenstate of $a^\dagger$.
The eigenstates of $a$, which are called "coherent states" are given by
$$
|\alpha \rangle = e^{-|\alpha|^2/2}\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}}|n\rangle \, .
$$
You can check easily by applying $a$ to $|\alpha \rangle$ that $|\alpha \rangle$ is an eigenstate of $a$.
A maximally classical state should have minimum and equally distributed uncertainty in $X$ and $P$. In other words the uncertain in $X$ should equal the uncertainty in $P$ and this uncertainty should be as small as possible. This leads to
$$
(X-\langle X\rangle)|\alpha\rangle=-i(P-\langle P\rangle)|\alpha\rangle
$$
or if we rearrange the equation
$$
\alpha|\alpha\rangle=\langle X+iP\rangle|\alpha\rangle=(X+iP)|\alpha\rangle=a|\alpha\rangle
$$
where $a=X+iP$ is the lowering operator. It is important to realize that $|\alpha\rangle$ is still a quantum state. As you have pointed out, $\langle X\rangle$ and $\langle P\rangle$ follow the classical trajectory, but if you calculate the variance of $X$ and the variance of $P$ you will find they are non-zero. The uncertainty principle must be satisfied.
Best Answer
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.