The following argument shows that if $C$ is essentially large (and locally small), then the presheaf $\Omega$ of sieves on $C$ is not small. Unfortunately, as Zhen points out, this does not show that $\mathcal{P}C$ does not have a subobject classifier, since its subobject classifier ought to be the presheaf of small sieves (i.e., sieves such that the corresponding presheaf is a small presheaf). In general, not all sieves are small, and in fact it seems rather difficult to construct any small sieves (exercise: construct a locally small category in which every object has a large set of sieves, but not a single nontrivial small sieve). It thus seems unlikely to me that this argument can be turned into a proof that $\mathcal{P}C$ does not have a subobject classifier without some strong additional hypotheses. Nevertheless, I'm keeping it as an answer in case you may find it somehow helpful (and because it is a nifty argument, even thought it doesn't prove what I hoped it would!).
Suppose that $\Omega$ is a small presheaf. Then every object $a\in C$ has only a small set of sieves, and there is a small set $S$ of objects such that for every $a\in C$ and every sieve $\sigma\in\Omega(a)$, there is an object $s\in S$, a map $f:a\to s$, and a sieve $\tau\in\Omega(s)$ such that $\sigma=f^*\tau$. In fact, if any such $\tau$ exists, then there is a canonical choice, namely $\tau=f_*\sigma$.
For any $a\in C$, let $\sigma_a$ be the sieve consisting of all maps $b\to a$ that do not have a right inverse (this is the second-largest sieve on $a$). We can thus find some $s_a\in S$ and $f_a:a\to s_a$ such that $\sigma_a=f_a^*f_{a*}\sigma_a$. Let's figure out explicitly what the condition $\sigma_a=f_a^*f_{a*}\sigma_a$ means. The sieve $f_{a*}\sigma_a$ consist of all maps of the form $f_ag$ where $g:b\to a$ does not have a right inverse. The sieve $f_a^*f_{a*}\sigma_a$ then consists of all maps $h$ such that $f_ah=f_ag$ for some $g$ that does not have a right inverse. The equation $\sigma_a=f_a^*f_{a*}\sigma_a$ holds iff the latter sieve does not contain the identity $1:a\to a$, or equivalently if every $g:a\to a$ such that $f_a=f_ag$ has a right inverse.
Since $S$ is small and $C$ is essentially large, we can find an essentially large set of objects $D\subseteq C$ such that all the objects $a\in D$ have the same associated object $s_a\in S$. That is, there is a single object $s\in S$ such that for each $a\in D$, there is a map $f_a:a\to s$ such that if $g:a\to a$ is such that $f_a=f_ag$, then $g$ has a right inverse. Now consider the sieves on $s$ generated by these maps $f_a$. Since $\Omega(s)$ is a small set, there is an essentially large set of objects $E\subseteq D$ such that the $f_a$ for $a\in E$ all generate the same sieve on $s$. That is, for any $a,b\in E$, there exist maps $g:a\to b$ and $h:b\to a$ such that $f_bg=f_a$ and $f_ah=f_b$. But then $f_ahg=f_a$, so $hg$ has a right inverse, and similarly $gh$ has a right inverse. In particular, $g$ and $h$ both have right inverses, and so $a$ and $b$ are retracts of each other. But for any object $a$, there are only an essentially small set of objects that are retracts of $a$, and this contradicts the essential largeness of $E$.
I'm not sure this constitute an answer to your question, but from what you are telling of your motivation it very well might be, and in any case it was too long to be a comment.
First - I don't think there is a good answer to your questions 1 and 2. The problem is that considering $\Omega$ as just a poset is a very "non-geometric" thing to do (in the sense of geometric logic) in fact this is somehow the typical non-geometric construction. So asking what its geometric properties are will probably not lead anywhere.
As such, I don't think there is any better description of this topos than the tautological one, that it classifies the ideals of $\Omega_E$, in the sense that a morphism $T \to \widehat{\Omega_E}$ classifies pairs of a morphism $f:T \to E$ and of an ideal of $f^*(\Omega_E)$. The non-geometric nature of $\Omega$ (when considered as a poset) means that $f^*(\Omega_E)$ can be pretty much anything, it has a comparison map to $\Omega_T$ of course, but that just means it has a distinguished subobject - so that not really an interesting observation.
However - there is another construction you can make that seems to fit what you are talking about more closely and produce a much nicer result. The thing is $\Omega$ might not be geometric when seen as a "set" or "poset", but it is geometric when seen with more structure - for example, it is obviously geometric when considered as a frame - but what might be more interesting to you given what you are saying - it is geometric when considered as a DCPO. (what I mean here is that if $f^\sharp$ denotes the left adjoint to $f_*$ acting on the categories of DCPO, then $f^\sharp \Omega_E = \Omega_T$). This is because $\Omega$ is the ind-completion of $\{0 < 1\}$.
So, if you really treat $\Omega$ as a DCPO you'll get much better results. This means you can't look at general presheaves on $\Omega$ though. But there is another way to get a topos out of a DCPO that will be geometric: You can use what Ivan Di liberty and myself have called the "Scott topos" (see here, and also Ivan Di Liberti's thesis and paper).
In Short, given a DCPO $P$, defines $S(P)$ as the category of (covariant) functor $P \to Set$ that preserves directed colimits. Then $S(P)$ is a (localic) topos, and you have a DCPO morphism from $P$ to the poset (DCPO) of points of $S(P)$. The Scott topos is a geometric construction on DCPO in the sense that the Scott topos $S(f^\sharp P)$ identifies with the pullback of the topos $S(P)$.
Moreover, in your present situation, as $\Omega$ is the ind-completion of {0<1} the Scott topos $S(\Omega)$ is exactly the Sierpinski topos.
Best Answer
Yes. The tightness axiom, $(a=b)\iff \neg (a\#b)$, implies $$\neg\neg(a=b) \iff \neg\neg\neg(a\#b) \iff \neg(a\#b) \iff (a=b).$$ Taking $b=\top$, we find that $\neg\neg a \iff a$ for all $a: \Omega$. This is the law of double negation, which is equivalent to excluded middle.