Laplace transform of $f\left(\frac{t - b}{a}\right)$
Your two alternate derivations of the Laplace transform (specifically the bilateral I'm referring to) of the signal $f((t - b)/a)$ seems right, both resulting in, the same , as a check.

$$
\mathcal{L}\{f((t - b)/a)\} = |a|e^{-bs}F(as)
$$

However, when you assume the signal $f(t) = e^t u(t)$ whose Laplace transform is
$$
F(s) = \frac{1}{s - 1}
$$

your conclusion is wrong
$$
\mathcal{L}\{f((t - b)/a)\} = |a|e^{-bs}F(as) \neq \frac{ae^{-(b/a)s}}{as - 1}
$$

which should be corrected as (according to your own derivation)
$$
\mathcal{L}\{f((t - b)/a)\} = |a|e^{-bs}\frac{1}{as - 1}
$$

So is this the problem you are referring to ?