I don't understand the question either. Does 'is there a lift?' mean any more or less than is image of the homomorphism contained in the subring?

The kernel of the map from $PSL_2(\mathbb{Z}_p)$ to $PSL_2(\mathbb{F}_p)$ is pro-$p$ so you can easily reduce your problem to the latter. Here $\mathbb{F}_p$ is the field with $p$ elements.

Is this mathoverflow.net/a/25231/345 an example?

Without rechecking the details I think you can argue that the centre must act on all the Ext groups by both the central character of $\pi_1$ and the central character of $\pi_2$. If the Ext group is Nonzero then the central characters must be equal. The argument is by 'functoriality'

The crucial thing for more general arguments along the lines I gave is the Artin-Wedderburn Theorem: en.wikipedia.org/wiki/Wedderburn%E2%80%93Artin_theorem. $\mathbb{F}G$ is a semisimple $\mathbb{F}$-algebra whenever $G$ is a finite group whose order is coprime to the characteristic of $\mathbb{F}$ (plus the case where the latter is $0$). As I hinted before the problem has two generalisations in the non-commutative setting depending on whether you wish 'ideal' to mean two-sided or one-sided ideals.

I guess that depends whether ideal means left ideal or two-sided ideal?

This kind of thing was proven by Cohen in jstor.org/stable/1990313 though it is rather more general, I think without explicitly containing the precise result, than you are asking about.

I think it was also excluded by the OP: 'Using a decomposition into disjoint cycles, one can simply compute what happens when multiplying by a transposition. This is not bad, but here the sign still feels like an ex machina sort of formula.'

Do you mean $A_i=A^i$ for each $i$ or are you fixing $i$ in some way?

What basis of the Lie algebra do you choose?

I'm not sure this question is tight enough though I suspect that the answer is no however it is clarified. Two initial problems: 'The PBW basis' isn't unique but depends on a choice of basis for $\mathfrak{g}$ and it isn't clear if you mean all irreducible submodules have such a basis or for some decomposition of the enveloping algebra as a direct sum of irreducibles all the summands have this property.

Actually maybe it is easy to prove that this infinite alternating group $A$ has no non-trivial finite dimensional representation since any non-trivial representation of the $A$ restricts to a non-trivial representation of $A_n$ for all $n\geq 3$ and the dimension of the smallest non-trivial representation of $A_n$ goes to infinity as $n$ goes to infinity.

I think the (simple) group of even finitely supported permutations of $\mathbb{N}$ (ie the 'union' of all finite alternating groups) may have this property though I don't know a proof of this fact.

I don't know the answer to question 1 but just wanted to make the trivial remark that you might as well assume that $\pi_1$ and $\pi_2$ are irreducible. I'd instinctively guess that the answer is no in general. If I wanted to think about it I'd start by considering $\pi_1$ and $\pi_2$ to be the irreducible $2$-dimensional representations of $S_3$ so $\pi_1\otimes \pi_2$ is uniquely decomposed as a direct sum of $3$ irreducible subreps. And then try to put inner products on $\pi_1$ and $\pi_2$ so that these are not mutually orthogonal in the tensor product.

I should say that this is essentially the same as Yemon Choi's answer which I hadn't seen until I posted but perhaps different enough to be worth leaving here.