Is this version of van der Waerden's Theorem consistent with ZFC? 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, provided that $\omega_1$ is the same in both models.
 A: I had pointed out earlier that the question as asked has a negative
answer, in light of the counterexample provided by my answer to
your previous question.
Theorem. There is a coloring of ordinals with two colors, such
that for any ordinals $\alpha$ and $\beta$, the ordinals
$\alpha+\beta\cdot\omega$ and $\alpha+\beta\cdot\omega^2$ get
different colors, using the usual ordinal arithmetic.

Update. But meanwhile, it was suggested on
your other question to use
natural (commutative) arithmetic
on the ordinals, and in that
case, the counterexample described above no longer works.
So let me provide a different counterexample here in the infinite case, using 
natural arithmetic. The principle (**) remains inconsistent with ZFC, even if one should use natural arithmetic. 
Theorem. For any infinite set $A\subset\mathbb{N}$, there is
coloring of the ordinals with two colors, such that there is no
$\alpha$ and $\beta$ such that $\alpha+\beta\cdot A$ is
monochromatic, using natural ordinal arithmetic.
Proof. First, we observe that the infinite case of van der
Waerden's theorem fails in the case of natural numbers:
Lemma. For any infinite $A\subset\mathbb{N}$, there is a
coloring of $\mathbb{N}$ with two colors, such that there are no
natural numbers $a$ and $b$ for which $a+b\cdot A$ is
monochromatic.
To prove the lemma, fix the set $A$, and now build up the coloring
by finite approximations. Since there are only countably many
pairs $(a,b)$, we may consider each of them in turn, and then
since $A$ is infinite, we may make sure to ensure that $a+b\cdot
A$ has a point of each color. So the lemma is proved.
Next, we lift this counterexample to the case of all ordinals as
follows. For any ordinal $\eta$, we may consider its unique
representation in Cantor normal form,
$$\eta=\omega^{\eta_n}+\cdots+\omega^{\eta_0}\quad\text{ where
}\eta_n\geq\cdots\geq\eta_0.$$ Let $T(\eta)$ be the number of
terms that appear in the Cantor normal form of $\eta$, which might
be called the trace of the ordinal. Note that $T(n)=n$ for any
finite number $n$, since $n=\omega^0+\cdots+\omega^0$.
Claim. $T:\text{Ord}\to\mathbb{N}$ is a homomorphism with
respect to natural addition and multiplication. That is,
 $$T(\alpha+\beta)=T(\alpha)+T(\beta)$$
 $$T(\alpha\cdot\beta)=T(\alpha)\cdot T(\beta),$$
 where on the left-hand side, we use natural arithmetic on the ordinals, and on the right hand side is the usual arithmetic on $\mathbb{N}$.
The claim can be proved by considering the definition of natural
sum and product. Basically, the natural sum $\alpha+\beta$ is
defined from the Cantor normal form by rearranging the terms from
the form of $\alpha$ and the form of $\beta$ so as to be in the
right order, and therefore it preserves the total number of terms.
Similarly, the natural product performs all possible products of
terms in the Cantor normal form (using distributivity) and
rearranges them into the right order, so that every cross term is
present, contributing one term to the Cantor normal form of the
result. (It would help for this to verify that
$\omega^\alpha\cdot\omega^\beta=\omega^{\alpha+\beta}$ using
natural sum and product, an issue I recall coming up on
MathOverflow previously.)
Now, fix the coloring $n\mapsto c(n)$ that is provided by the
lemma for the given infinite set $A\subset\mathbb{N}$. Extend this
coloring to all ordinals by defining $c(\eta)=c(T(\eta))$. That
is, we color an ordinal $\eta$ according to the color that the
trace $T(\eta)$ would get via $c$.
I claim finally that for any ordinals $\alpha$ and $\beta$, the
collection of ordinals given by $\alpha+\beta\cdot A$ is not
monochromatic, using natural arithmetic. To see this, observe that
the color of $\alpha+\beta\cdot n$ is the same as the color of its
trace in the natural numbers $T(\alpha+\beta\cdot
n))=T(\alpha)+T(\beta)\cdot n$, using the homomorphism properties of $T$, and so if $\alpha+\beta\cdot A$
were monochromatic, then so would be $a+b\cdot A$, where
$a=T(\alpha)$ and $b=T(\beta)$. But by the choice of the coloring
on the natural numbers, this is impossible. QED
