The answer is yes, because every forcing notion is equivalent to a forcing notion with finite predecessors.

**Theorem.** Every forcing notion is forcing equivalent to a forcing notion with finite conditions, a family of finite sets ordered by (reverse) inclusion. All such forcing notions have the finite-predecessor condition mentioned in the question.

**Proof.** Consider any forcing notion $\newcommand\P{\mathbb{P}}\P$. Let $\P^*$ be the forcing notion consisting of the finite pointed subsets of $\P$, that is, finite sets $a\subset\P$ such that $a$ has a least element. We order $\P^*$ by $a\leq b$ if and only if $b\subseteq a$. Since a finite set has only finitely many subsets, this will ensure that $\P^*$ has the finite-predecessor condition.

It is easy to see that $\P^*$ projects to $\P$ by mapping every pointed set to its point. So forcing with $\P^*$ adds a generic for $\P$. 

Conversely, I claim that forcing with $\P$ adds a generic for $\P^*$. Assume that $G\subset\P$ is $V$-generic, and let $G^*$ consist of the finite pointed subsets of $G$. This is a filter in $\P^*$, since $G$ is a filter in $\P$. This is $V$-generic for $\P^*$, since if $D^*\subset\P^*$ is dense, then let $D$ consist of the least points of any pointed set in $D^*$ — this will be dense in $\P$. So there is some $q\in D$ which is least in some $b\in D^*$ such that $q\in G$. So $b\subset G$ and thus $b\in G^*$ and so $G^*$ meets $D^*$. So $G^*$ is $V$-generic for $\P^*$. 

Since $G$ and $G^*$ are easily constructed from each other, we have $V[G]=V[G^*]$, and so these forcing notions are forcing equivalent. $\Box$

Meanwhile, as Monroe mentions [in the comments](https://mathoverflow.net/questions/479025/example-of-a-forcing-notion-with-finite-predecessor-condition-that-does-not-add/479045#comment1246527_479045), a drawback of this observation is that the forcing notion $\P^*$ is not generally separative, since pointed sets in $\P$ can use the same least point but be inclusion incomparable, so they will be incomparable in the $\P^*$ order but compatible with exactly the same conditions, violating separativity. I would view it as a natural version of the question to inquire about separative posets with the finite-predecessor property.