There are very concrete examples of matrix groups $G$ with this property. See for example the paper: Yves de Cornulier, Finitely presentable, non-Hopfian groups with Kazhdan’s property (T) and infinite outer automorphism group, Proc. Amer. Math. Soc. 135 (2007), no. 4, 951-959.

The second one cannot be a zero-divisor in $\mathbb CG$. This is a standard argument looking at the action of $\mathbb CG$ on $\ell^2(G)$. If some non-zero vector $\xi \in \ell^2(G)$ is annihilated by $4+x+x^{-1}+y+y^{-1}$ one can show using computations with scalar products that $-x\xi = \xi$. But that is impossible if $x$ has infinite order. For the other two, I do not think that it is known if they can be zero-divisors or not.

If $(A_1,A_2) \in SL(2,\mathbb C)^2$ does not have an invariant subspace in $\mathbb C^2$ (this is the generic case), then its simultanous conjugacy class is determined by the triple $(tr(A_1), tr(A_2), tr(A_1A_2)) \in \mathbb C^3$. A natural guess would be that the only constraint is that this triple lies in $\mathbb Z^3$, but I do not know if that is true.

Have a look at Theorem 7.8 in (W. Woess. Random walks on infinite graphs and groups. Cambridge Tracts in Mathematics Vol. 138, Cambridge University Press, 2000.) and/or the remarks after Excercise 14.32 in the notes of Gabor Pete, math.bme.hu/~gabor/PGG.pdf. I do not know about $l^p(G)$, but I would guess it is the same.

The answer is always negative for infinite groups. This is a property of the random walk on the group. I remember that a result in the book of Woess implies that $\tau(S^n)/\|S\|^n$ converges zero. However, the weak limit is (or contains) the spectral projection at the point $\|S\|$.

I edited the last sentence to clarify the meaning.

Ok, I do not understand the proof either (unless it is the argument from above). The confusion might arise from the fact that many representations are "isomorphic", but this is not distinguished from being "identical" in the Wikipedia article.

Yes, exactly. No proper power of $\chi^2$ is golden.

I do not think that there is a problem with the argument. The Schröder-Bernstein theorem is almost a triviality once you get the right picture. Just draw a bipartite graph with edges coming from $f$ and $g$ and consider the connected components -- either its a cycle, a two-sided line or a one-sided line. Hence, for each component, there is a trivial bijection (supported on the bipartite graph). All other proofs of generalizations follow this simple recipe, maybe involving some partition process to get started.