The answer is yes, and it's fairly elementary. By the usual 2-descent, the curve $C$ gives a class $c$ in $H^1(\mathbb{Q},E[2])$, where $E$ is the Jacobian you wrote down. As you vary $d$, the groups $H^1(\mathbb{Q},E_d[2])$ are canonically isomorphic, and $c$ is also the class of $C_d$. To answer your question, you should condition on the possibility that $c$ "comes from" the 2-torsion subgroup of $E$ via $E(\mathbb{Q})/2E(\mathbb{Q}) \to H^1(\mathbb{Q},E[2])$. I don't know what the conditions on $a,b$ for this to happen are, but it won't matter.

If $C$ doesn't come from 2-torsion, and if $C_d$ has a rational point, then automatically $E_d$ has rank at least one (by descent). And as Alex B. says, it's easy to see that there are infinitely many squarefree $d$ for which this happens (his argument doesn't quite show this, but it's not hard to fiddle with congruence conditions to produce infinitely many squareclasses).

If $C$ does come from 2-torsion, then $C_d$ has a rational point for all $d$, but there is no guarantee that the rank is at least 1. So your question reduces to asking about the ranks of the twists of $E$. But now just apply the argument from the previous paragraph to some other class in $H^1(\mathbb{Q},E[2])$ represented by a binary quartic form (there are infinitely many of these).

In particular, you don't need to know anything about Sha or to invoke Daniel Kane's work.

The answer is yes, and it's fairly elementary. By the usual 2-descent, the curve $C$ gives a class $c$ in $H^1(\mathbb{Q},E[2])$, where $E$ is the Jacobian you wrote down. As you vary $d$, the groups $H^1(\mathbb{Q},E_d[2])$ are canonically isomorphic, and $c$ is also the class of $C_d$. To answer your question, you should condition on the possibility that $c$ "comes from" the 2-torsion subgroup of $E$ via $E(\mathbb{Q})/2E(\mathbb{Q}) \to H^1(\mathbb{Q},E[2])$. I don't know what the conditions on $a,b$ for this to happen are, but it won't matter.

If $C$ doesn't come from 2-torsion, and if $C_d$ has a rational point, then automatically $E_d$ has rank at least one (by descent). And as Alex B. says, it's easy to see that there are infinitely many squarefree $d$ for which this happens (his argument doesn't quite show this, but it's not hard to fiddle with congruence conditions to produce infinitely many squareclasses).

If $C$ does come from 2-torsion, then $C_d$ has a rational point for all $d$, but there is no guarantee that the rank is at least 1. So your question reduces to asking about the ranks of the twists of $E$. But now just apply the argument from the previous paragraph to some other class in $H^1(\mathbb{Q},E[2])$ represented by a binary quartic form (there are infinitely many of these).