You may have already tried, but doesn't proposition 3.2.3 of Michel-Venkatesh's subconvexity paper give what you want?

For your $x$ we have $f(1)\neq 0$. Because of continuity there exists an open ball around $1$ which is contained in the support of $f$.

@Kimball I meant that you can recover $f(g)$ for all $g\in G$ by suitably translating $f$ inside the Plancherel integrand by a regular representation.

@Kimball I think the OP is calling $tr(\pi(f))$ to be Fourier transform of $f$ as an analogue of one dimensional representation (as if $tr\pi$ is a character). Also I am not sure what you meant by "not invertible". One can get back $f$ by a Plancherel integral: $$f(1)=\int_{\hat{G}}tr(\pi(f))d\mu(\pi).$$

....This modular domain has one cusp, which can be checked via strong approximation. You can talk about the Maass forms on this space, for example, a real analytic Eisenstein series would be $$E_s(P):=\sum_{\gamma\in\Gamma_\infty\backslash\Gamma}r(\gamma P)^s.$$

@johnmangual The hyperbolic $3$ space can be realized as an upper half plane model by a subset of Hamiltonian quaternions having zero real part: $$\{P:=z+rj\mid z\in \mathbb{C}, r\in \mathbb{R}^+\}.$$ $SL_2(\mathbb{C})$ acts transitively by generalized Mobius transformation: $$\begin{pmatrix}a&b\\c&d\end{pmatrix}P=(aP+b)(cP+d)^{-1}.$$ As quaternions are not commutative $\frac{aP+b}{cP+d}$ does not make sense. Then the space $$SL_2(\mathbb{Z}[i])\backslash \mathbb{H}^3 = SL_2(\mathbb{Z}[i])\backslash SL_2(\mathbb{C})/SU(2).$$

@Corbennick It should be $PSL_2(\mathbb{C})$.

It seems decaying. So two $e^{-\pi t/2}$ coming from two $K$-Bessel functions should cancel $\sinh(\pi t).$ Could you give me any explicit bound in $t$, or some reference? Thanks.