The following theorem seems to express how the various definability witness properties are connected with each other and with $V=\text{HOD}$.
Theorem. The following are equivalent in any model $M$ of ZF:
$M$ is a model of $\text{ZFC}+\text{V}=\text{HOD}$.
$M$ has a definable well-ordering of the universe.
Every definable nonempty set in $M$ has a definable element.
Every definable nonempty set in $M$ has an ordinal-definable element.
Every $(\Sigma_2\wedge\Pi_2)$-definable nonempty set in $M$ has an ordinal-definable element.
Every ordinal-definable nonempty set in $M$ has an ordinal-definable element.
Proof. ($1\to 2$) The usual HOD order is a definable well-ordering of the universe.
($2\to 3$) Select the least element with respect to the definable order, as in Bjorn's answer.
($3\to 4$) Immediate.
($4\to 5$) Immediate.
($4\to 1$) If $M$ thinks there is a non-OD set, then the set $A$ of all non-OD sets in $M$ of minimal rank is a definable nonempty set in $M$ with no ordinal-definable elements.
($5\to 1$) For the stronger implication — thanks to François in the comments! — we analyze the complexity of the definition of $A$ in the previous case. The main observation is that $x\in\text{OD}$ is $\Sigma_2$, as every instance of definability is witnessed by reflection inside some $V_\theta$. The $\Sigma_2$ assertions are precisely the semi-local properties, those that can be verified in some rank initial segment of the universe $V_\theta$.
If $V\neq\text{HOD}$, then the set $A$ of minimal-rank non-OD sets is characterized by the following properties: $A$ is not empty; all elements of $A$ have the same rank; every element of $A$ is not in OD; every set of rank less than an element of $A$ is in OD; every set not in $A$, but of the same rank as an element of $A$, is in OD. Each of these properties is either $\Sigma_2$ or $\Pi_2$, making the set $A$ to be $\Sigma_2\wedge\Pi_2$-definable. Specifically, the first two requirements are $\Sigma_2$, being witnessed in a rank-initial segment of the universe; the third is $\Pi_2$; the fourth and fifth are both $\Sigma_2$, since they are true just in case there is a large $V_\theta$ which believes them to be true. Thus, the definition of $A$ has complexity $\Sigma_2\wedge\Pi_2$, which allows the implication ($5\to 1$), since $A$ has no ordinal-definable members.
($1\to 6$) Immediate, since under statement $1$, every set in $M$ is ordinal-definable in $M$.
($6\to 4$) Immediate. QED
Conclusion. Thus, case (1) of the question occurs in exactly the models of $V=\text{HOD}$ that are not pointwise definable. There are such models, if ZFC is consistent, since one may take any uncountable model of $\text{ZFC}+V=\text{HOD}$.
Meanwhile, case (2) of the question does not occur at all, since if a set has sets that are not ordinal-definable, then it will have a definable set with no ordinal-definable members, namely, the set of all non-OD sets of minimal rank, as in the implication of statement 4 to statement 1.