I found the answer, at least for some of the cases that I thought were intractable, including my examples. The main tool is the disentangling theorem for the relevant group.

__Example 1__: the transformation is in $SU(2)$, so we need the $SU(2)$ disentangling theorem:
$$
\exp(z J_+-z^* J_-) = \exp(\tau_+ J_+)\exp(\tau_ 0J_0)\exp(\tau_-J_-)
$$
Here $\{J_0,J_\pm\}$ satisfy the $su(2)$ algebra relations: $[J_\pm,J_0]=\mp J_\pm,\ [J_-,J_+]=-2J_0$, and if $z=re^{i\phi}$, then $\tau_\pm=\pm e^{\pm i\phi}\tan(r),\tau_0=2\log\sec(r)$.
If we apply this to the beamsplitter transformation in the form $U(\theta)=\exp[\theta(a^\dagger b-ab^\dagger)]$, we obtain
$$
U(\theta) = \exp(\tan(\theta)a^\dagger b)\exp[\log\sec(\theta)(a^\dagger a-b^\dagger b)]\exp(-\tan(\theta)a b^\dagger)
$$
which has no truncation issues.

__Example 2__: the transformation is in $SU(1,1)$, so we need the $SU(1,1)$ disentangling theorem:
$$
\exp(z K_+-z^* K_-) = \exp(\sigma_+ K_+)\exp(\sigma_0K_0)\exp(\sigma_-K_-)
$$
Here $\{K_0,K_\pm\}$ satisfy the $su(1,1)$ algebra relations: $[K_\pm,K_0]=\mp K_\pm,\ [K_-,K_+]=2K_0$, and if $z=re^{i\phi}$, then $\sigma_\pm=\pm e^{\pm i\phi}\tanh(r),\sigma_0=-2\log\cosh(r)$. If we apply this to the squeezing operator $S(z)=\exp[z \frac{{a^\dagger}^2}{2}-z^* \frac{{a}^2}{2}]$, we obtain
$$
S(z) = \exp\left(e^{i\phi}\tanh(|z|)\frac{{a^\dagger}^2}{2}\right)\exp\left(-\log\cosh(|z|)(a^\dagger a+\frac{1}{2})\right)\exp\left(-e^{-i\phi}\tanh(|z|)\frac{{a}^2}{2}\right)
$$
which has no truncation issues.

__General case__: in general it might be impossible to have a suitable disentangling theorem, but results like [this][1] might help approximating it (see eq. (30)-(35) for the two examples above).

I hope this will help those who will stumble upon my same issue.


  [1]: http://aip.scitation.org/doi/10.1063/1.3413923