I am looking for infinite families of prime knots that have all Alexander polynomial equals to 1. I wrote "families" (and not "family") since perhaps there are different constructions out there.
Moreover, is there such a family for which all knots are of genus one (just as the Whitehead doubles)?
The P(p,q,r) pretzel knots with p, q, r odd integers have a genus 1 Seifert surface that just consists of two disks connected with three twisted ribbons, with p, q, r twists, respectively. So they have a Seifert matrix $A = \left(\begin{matrix}p+q & q+1 \\ q-1 & q+r\end{matrix}\right)/2$.
From this one computes that the determinant of P(p,q,r) is $|\det(A + A^{\top})| =pq+qr+rp$. For genus 1 knots, the Alexander polynomial is 1 if and only if the determinant is 1. This gives you plenty of examples of genus 1 knots with Alexander polynomial 1, e.g. P(-3,5,7).
More generally, given any knot $K$ with Alexander polynomial 1 and a Seifert surface $\Sigma$, you can construct further knots of the same genus as $K$ simply by tying a knot into a band of $\Sigma$ without changing the framing of the core curve of the band (this is often called "infection").
Whitehead doubles are one such family of genus one knots, see this MO answer
Another family of examples is given by the "generalised Kinoshita-Terasaka" knots, here is a picture from Lickorish' "An introduction to Knot Theory".
Here $d$ is assumed to be even. Of course, this is not a family of genus 1 knots.
