Fourier Transform of a signal using direct integration and properties
Your first solution using the properties of the Fourier transform is correct. Your second solution is wrong, because you forgot to include the unit step function. Your function $g(t)$ should be defined by

$$g(t)=e^{-t}u(t)\tag{1}$$

which gives for $g(2t-1)$

$$g(2t-1)=e^{-(2t-1)}u(2t-1)=e^{-(2t-1)}u\left(t-\frac12\right)\tag{2}$$

Consequently, the Fourier integral is given by

$$\mathcal{F}\{g(2t-1)\}=\int_{\frac12}^{\infty}e^{-(2t-1)}e^{-j\omega t}dt\tag{3}$$

I'm sure you can prove now that the result of $(3)$ equals the result you got from using the properties of the Fourier transform.