One way to phrase van der Waerden's Theorem is:

For every finite coloring of $\mathbb N$ and every finite $F \subseteq \mathbb N$, there exist $a,b \in \mathbb N$ such that $a + b \cdot F$ is monochromatic.

In this question I ask about one possible extension of this theorem to $\omega_1$, which I call $(*)$. In this question I'm asking about another possible extension:

$(**)$ For every finite coloring of $\omega_1$ and every countable $C \subseteq \omega_1$, there exist $\alpha,\beta \in \omega_1$ such that $\alpha + \beta \cdot C$ is monochromatic.

It's fairly easy to show that $(**)$ is equivalent to:

$(**)'$ For every finite coloring of $\omega_1$ and every initial segment $I$ of $\omega_1$, there exist $\alpha,\beta \in \omega_1$ such that $\alpha + \beta \cdot I$ is monochromatic.

Question: Is $(**)$ consistent with ZFC?

Remark 1: Usually the elements of $\omega_1$ are also used to represent the initial segments of $\omega_1$ (i.e., every ordinal is the set of its predecessors). That seems confusing here, which is why I'm using $I$.

Remark 2: The negation of $(**)$ is consistent with ZFC. To see this, force with finite partial functions $\omega_1 \rightarrow 2$ (i.e., add $\aleph_1$ Cohen reals). Interpret the generic object as a $2$-coloring of $\omega_1$. A simple argument using dense sets shows that for every infinite initial segment $I$ and every $\alpha, \beta$, $\alpha + \beta \cdot I$ cannot be monochromatic. In fact, since this argument mentions only $\aleph_1$ dense sets and the forcing is ccc, the same argument shows that $(**)$ is incompatible with MA.

Remark 3: If some coloring gives a counterexample to $(**)$ in one model of set theory, it continues to give a counterexample in any larger model. In other words, $(**)$ is downward-absolute. Thus, to give a negative answer to my question, it's enough to show that $(**)$ is inconsistent with $V = L$. Conversely, a positive answer to my question means that $(**)$ is consistent with $V = L$. This means that (most likely) the only way to get a positive answer is to show that $V = L$ implies $(**)$. In other words, what I'm really asking is:

Alternate version of question: Does $(**)$ hold in $L$?