Is this an equivalent condition for algebraic posets?
An example where $A(x)$ is empty is the set of real numbers $\mathbb{R}$ with the usual ordering. It has no compact elements at all.

If we assume the second condition then $A(x)$ cannot be empty: if $A(x) = \emptyset$ then by the second condition $x$ is the empty join, therefore the least element of $L$, which is compact, therefore $x \in A(x) = \emptyset$, a contradiction.

Your proposal to replace the first condition with non-emptyness does not work. Consider the poset $L$ which consists of two copies of $\mathbb{N}$ and $\infty$, where we write $\iota_1(n)$ and $\iota_2(n)$ for the two copies of $n$, ordered by:


$\iota_1(m) \leq \iota_1(n) \iff m \leq n$
$\iota_2(m) \leq \iota_2(n) \iff m \leq n$
$x \leq \infty$ for all $x$.


In words, we have two incomparable chains with a common supremum. All elements are compact except $\infty$. Now:


${\downarrow}x \cap K(L) \neq \emptyset$, obviously.
$x = \bigvee ({\downarrow}x \cap K(L))$, obviously.
The set ${\downarrow}\infty \cap K(L) = \mathbb{N} + \mathbb{N}$ is not directed.