Here is some insight for you.
Consider a partially ordered set $S$ which has all meets. We will prove that $S$ has all joins.
Let us consider some indexed collection $\{s_i \in S\}_{i \in I}$. Let $J = \{x \in S \mid \forall i \in I (s_i \leq x)\}$. I claim that $\bigwedge J$ is the join of $s$.
Indeed, we first note that for all $i$, for all $x \in J$, $s_i \leq x$ (by the very definition of $J$). Therefore, $s_i \leq \bigwedge J$.
Now suppose that for all $i \in I$, $s_i \leq x$. Then $x \in J$, so $\bigwedge J \leq x$. This completes the proof. $\square$
Your definition also works. Let's work through it.
First, we must show that $Q := \{ \bigvee X \mid X \subseteq \bigcup\limits_{\alpha \in \mathcal{J}} I_\alpha$ and $\bigvee X$ exists$\}$ is in fact a $c$-ideal.
As you have shown, (1) is trivial.
For (2), suppose that $X \subseteq \bigcup\limits_{\alpha \in \mathcal{J}} I_\alpha$ and $\bigvee X$ exists, and that $a \leq \bigvee X$. Then $a = a \land \bigvee X = \bigvee \{a \land x \mid x \in X\}$ (using the fact that $a \land$ is the left adjoint to $a \implies$ and therefore preserves any colimits that do exist). Since each $I_\alpha$ is downward closed, if $x \in I_\alpha$ then $a \land x \in I_\alpha$. And thus, we see that $\{a \land x \mid x \in X\} \subseteq \bigcup\limits_{\alpha \in \mathcal{J}} I_\alpha$. Thus, $a \in Q$.
For (3), we need to get a big cleverer if we wish to avoid choice. Given $a \in Q$, let $f(a) = \{w \in \bigcup\limits_{\alpha \in \mathcal{J}} I_\alpha \mid w \leq a\}$; then clearly, $a = \bigvee f(a)$.
Now suppose we have some family $\{x_k \in Q\}_{k \in K}$, and that $\bigvee\limits_{k \in K} x_k$ exists. Then in particular, we have $\bigvee\limits_{k \in K} x_k = \bigvee\limits_{k \in K} \bigvee f(x_k) = \bigvee \bigcup \limits_{k \in K} f(x_k)$. Note that $f(x_k) \subseteq \bigcup\limits_{\alpha \in \mathcal{J}} I_\alpha$ by definition, so $\bigcup \limits_{k \in K} f(x_k) \subseteq \bigcup\limits_{\alpha \in \mathcal{J}} I_\alpha$. Therefore, $\bigvee\limits_{k \in K} x_k = \bigvee \bigcup \limits_{k \in K} f(x_k) \in Q$.
Now, let's go about proving that $Q$ is in fact the join of the $I_\alpha$.
Clearly, we see that if $x \in I_\alpha$, then $x = \bigvee \{x\} \in Q$. So $I_\alpha \subseteq Q$ for all $\alpha$.
Now suppose that we had some $y$ such that for all $\alpha$, $I_\alpha \subseteq y$. Then condition 3 ensures that $Q \subseteq y$.
Best Answer
No. Suppose $\mathbf{A}$ is the lattice of open sets in some topological space $X$. Then for each $x\in X$, the filter $F$ of open sets containing $x$ is prime, and the quotient $\mathbf{A}/F$ can be identified with the lattice of germs of open sets at $x$ (i.e., the colimit of the lattices of open subsets of $U$ where $U$ ranges over all neighborhoods of $x$).
In particular, let $X=\mathbb{N}\cup\{\infty\}$ and $x=\infty$. Then the elements of $\mathbf{A}/F$ besides $1$ can be identified with the quotient of the Boolean algebra $\mathcal{P}(\mathbb{N})$ by the ideal of finite sets (an element of $\mathbf{A}/F$ besides $1$ is a germ of an open set that doesn't contain $\infty$, and that's just a subset of $\mathbb{N}$; two such subsets agree on a neighborhood of $\infty$ iff they differ by a finite set). That is, $\mathbf{A}/F$ is the lattice $\mathcal{P}(\mathbb{N})/\mathrm{fin}$ with a greatest element added on top. Since $\mathcal{P}(\mathbb{N})/\mathrm{fin}$ is not complete, neither is $\mathbf{A}/F$.
(Note that it is easier to come up with prime filters that are not closed under arbitrary joins--for instance, take a nonprincipal ultrafilter in $\mathcal{P}(\mathbb{N})$, or consider the case where $\mathbf{A}$ is totally ordered. However, in those easy examples the quotient $\mathbf{A}/F$ is actually still complete!)