Denote by $\sigma$ the involution on the abelian surface $A$ and set $X:=A/\sigma$. The eigenvalues of the action of $\sigma$ on $H^0(\Omega^1_A)$ are $+1$ and $-1$.
So $\sigma$ has no isolated fixed points and it follows that $X$ is smooth with $h^1(\mathcal O_X)=1$
There are two possibilities:

1) $\sigma$ has no fixed point. Then the quotient map $A\to X$ is \'etale and $X$ is a bielliptic surface almost by definition. 

2) $\sigma$ has a fixed point. Then, up to a translation,  we may assume that $\Sigma$ is an endomorphism. The fixed locus is a union of elliptic curves and $X$ is ruled. This can be seen either by classification of surfaces or observing that $A$ contains a family of translates of elliptic curves on which $\sigma$ acts as multiplcation by $-1$.