A function is a bijection iff it has an inverse, so one way to prove the statement is to show that the inverse function to $\lambda_g$, $(\lambda_g)^{-1} = \lambda_{g^{-1}}$.
You could also just prove that $\lambda_g$ is surjective and injective directly, which isn't too hard to do.
To show that $\lambda_g$ is surjective, take $h \in G$. Then $\lambda_g(g^{-1}h) = gg^{-1}h = h$. $g^{-1}h \in G$ so we're done.
To show that $\lambda_g $ is injective, suppose $\lambda_g(h) = \lambda_g(h')$. Then $gh = gh'$ so by multiplying both sides on the left by $g^{-1}$ we have $h = h'$, so $\lambda_g$ is injective.
Edit: Also, as far as I'm aware, what you call the left regular representation is called the left regular action. It may just be two ways of saying the same thing, but as you had (??) in your post I thought I'd add this in anyway.