You want $S$ to be an isomorphism from $V$ onto a subspace, say $X$. (You also want $X$ to be complemented, but that is irrelevant for the answer.) This implies that $S^{''}$ is an isomorphism from $V$ onto

$X^{\perp\perp} \subset V^{''}$. So no operator like you want exists on any non reflexive Banach space.

You want $S$ to be an isomorphism from $V$ onto a subspace, say $X$. (You also want $X$ to be complemented, but that is irrelevant for the answer.) This implies that $S^{''}$ is an isomorphism from $V^{''}$ onto

$X^{\perp\perp} \subset V^{''}$. So no operator like you want exists on any non reflexive Banach space.