With respect to ordinal-definability, the answer is no, this is not consistent. The reason is that if there is any non-OD set, then there is such a set of minimal rank. Thus, the set of all non-OD sets of that rank is a nonempty (paremeter-free) definable set, with no ordinal-definable member. Thus, the assertion "every nonempty ordinal-definable set has an ordinal-definable member" is simply provably equivalent to $V=\text{HOD}$.
Theorem. The following are equivalent in any model $M$ of ZF:
$M\models \text{V}=\text{HOD}$.
Every nonempty ordinal-definable element of $M$ has an ordinal-definable element.
Every nonempty definable (without parameters) element of $M$ contains an ordinal-definable definable element.
Proof. Clearly 1 implies 2 and 2 implies 3. For 3 implies 1, note that if $V=\text{HOD}$ fails, then the set of rank-minimal sets not in $\text{HOD}$ is a nonempty definable set containing no ordinal definable element, violating 3. QED