Let X,Y be Banach spaces and S a bounded operator $S: Y' \rightarrow X'$ where $'$ denotes the dual space or the adjoint operator depending on what it is on.
Then
$$ \exists \ \ T \in B(X,Y) : T'=S \iff S'(\hat{X}) \subset \hat{Y} $$
where $\hat{X}$ is the image of $X$ under the canonical isometry $\phi:X \rightarrow X''$
$$\phi(x)x'= x'(x) \ \ \ \ \ \ \ \ \ \ \ \ \forall x\in X, x'\in X'$$
I fail to see how the completeness assumption is used. I imagine it will be useful since it gives us that both $\hat{X}$ and $\hat{Y}$ are closed.
This is what I thought:
Assume $ S'(\hat{X}) \subset \hat{Y}$ then for each $x$ there is an element of $Y$ (call it $Tx$) such that $S'(\hat{x})= \widehat{Tx}$
then let $y'\in Y'$,
$S( y'(x))= \hat{x}(S(y'))=S'(\hat{x})(y')= (\widehat{Tx})y'=y'(T(x))=(T'y')x$
therefore $S=T'$ (assuming such $T'$ exists, but it is not too hard to show that T is a bounded linear operator)
I think that for the converse the argument is essentially the same... what am I missing?
Completeness is not needed. People often state results for Banach spaces because this is the context they are working in; even though the result may be true for general normed linear spaces. Besides, "Banach" is shorter than "normed linear".
For any $S :Y'\to X'$ there is one, and only one, candidate for the pre-adjoint of $S$: namely, the restriction of $S'$ to the image of $X$ in $X''$. The only issue is whether this candidate takes values in $Y$. The diagram-chasing part of solution does not involve any topology, in particular does not rely on completeness.
(Tangential: here is a nice example of a surjective isometry without a pre-adjoint.)