$$\int_{-\infty}^{\infty} g(t)dt = \int_{0}^{\infty} \frac{\lambda^{\alpha}}{\Gamma(\alpha)} t^{\alpha - 1} e^{-\lambda t} dt = \frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\infty} (\lambda t)^{\alpha - 1} e^{-\lambda t} dt$$
Set $u = \lambda t$:
$$= \frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\infty} u^{\alpha - 1} e^{-u} du = \frac{\lambda}{\Gamma(\alpha)} {\Gamma(\alpha)} = \lambda$$
I've tried this proof several times and always end up with an extra $\lambda$. Each step seems pretty straightforward so I can't find what I'm overlooking.