We claim that both statements are equivalent to $\angle A = 60^\circ$.

Firstly,
$A, D', E'$ are collinear
$$\Leftrightarrow \angle BAD' + \angle CAE' = \angle BAC$$
$$\Leftrightarrow \angle BAD + \angle CAE = \angle BAC$$
$$\Leftrightarrow \frac{C}{2} + \frac{B}{2} = A$$
$$\Leftrightarrow 90^\circ - \frac{A}{2} = A$$
$$\Leftrightarrow A = 60^\circ.$$

Secondly, it is well-known that the reflection of $H$ in $AB$ lies on $(ABC)$. Therefore, $A, B, H, D'$ are concyclic. Also, note that $D, D', O$ are collinear since all of them lie on the perpendicular bisector of $AB$. It follows that
$$\begin{align*} \angle DD'H &= \angle DD'B + \angle BD'H \\ &= (90^\circ - \angle ABD') + \angle BAH \\ &= (90^\circ - \angle ABD) + \angle BAH \\ &= \left(90^\circ - \frac{C}{2}\right) + 90^\circ - B \\ &= A + \frac{C}{2}. \end{align*}$$

Similarly, $C, A, H, E'$ are concyclic, and we have
$$\begin{align*} \angle OE'H &= \angle CE'H - \angle CE'E \\ &= (180^\circ - \angle CAH) - (90^\circ - \angle ACE) \\ &= (90^\circ + C) - \left(90^\circ - \frac{B}{2}\right) \\ &= C + \frac{B}{2}. \end{align*}$$
Now,
$O, H, D', E'$ are concyclic
$$\begin{align*} \angle DD'H &= \angle OE'H \\ A + \frac{C}{2} &= C + \frac{B}{2} \\ A &= 90^\circ - \frac{A}{2} \\ A &= 60^\circ. \end{align*}$$

Therefore, the two statements are equivalent.