After a bit of thought, I think I have a pretty good set for $\mathcal E^{fin}$ and a pretty good class for $\mathcal E$.

**Proposition:** Let $L$ be a finite poset. Then $L$ is a lattice if and only if $L$ has the right lifting property with respect to the following set $\mathcal E^{fin}$ of embeddings:

 1. $\emptyset \to 1$

 2. $2 \to 2^\triangleleft$ and $2 \to 2^\triangleright$ where $2$ is the discrete poset with two elements and these morphisms add a cone and cocone respectively.

 3. $\newcommand{\bbowtie}{\bowtie\mkern-17mu\bullet\mkern17mu}$ $\bowtie \to \bbowtie$ where $\bbowtie$ is the 5-element poset $x,y < a < p,q$ and $\bowtie$ is the full sub-poset on $x,y,p,q$.

**Proof:**
It suffices to show that $L$ is a meet-semilattice, i.e. that for every finite set $S \subseteq L$, the poset $L \downarrow S$ of elements under $S$ has a top element. It suffices to consider the cases (i) when $S$ is empty and (ii) when $S$ has two elements. Moreover, every finite directed poset has a top element, so it will suffice to show that (i) $L$ is directed and (ii) for every $p, q \in L$, the poset $L \downarrow \{p,q\}$ is directed. (i) follows from (1) and the second part of (2). (ii) follows from the first part of (2) and (3).

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Analogously, using $\infty$-directedness in place of directedness, we have


**Proposition:** Let $L$ be a poset. Then $L$ is a complete lattice if and only if $L$ has the right lifting property with respect to the following class $\mathcal E$ of embeddings:

 1. $\emptyset \to 1$

 2. $S \to S^\triangleleft$ and $S \to S^\triangleright$ for each discrete poset $S$ (i.e. we add a top element and a bottom element, respectively, to $S$)

 3. $\newcommand{\bbowtie}{\bowtie\mkern-17mu\bullet\mkern17mu}$ $\bowtie_{S,S} \to \bbowtie_{S,S}$ for each set $S$, where $\bbowtie_{S,T}$ is the poset whose under lying set is $S \amalg T \amalg \{a\}$, with $S < a < T$, and $\bowtie_{S,T}$ is the full subposet on $S \amalg T$.