The answer is always "no". By classification, a bielliptic surface over $\mathbb C$ has the form $(E\times F)/G$ where $E,F$ are elliptic curves, $G=\mathbb Z_n=(g)\subset G=\subset Aut(E,0)$ is an abelian group acting on $E\times F$ by $g.(x,y)= (g.x,y+b)$, with $nb=0$. So $G$ acts as a complex multiplication multiplications on $E$ and as a translation by translations on $F$. ($G$ is not necessarily cyclic as Tuan correctly points out.)
($X$ maps to an elliptic curve $F/\mathbb Z_n$ F/G$ and every fiber is isomorphic to an elliptic curve $E$, hence the name bielliptic.)
Then $F$ acts on $E\times F$ by $(x,y)\mapsto (x,y+f)$, and this action commutes with the $G$-action. Thus, $F\subset Aut^0(X)$. As $F$ is a projective variety, $Aut^0(X)$ is not affine.