= Solution
\b[Yes.] Use the sequence $x_j=2\cdot10^{j-1}$, $j\geq1$. Every nonempty finite sum has only digits $0$ and $2$ in decimal notation, with leading digit $2$, so none is a power of $10$, including $10^0=1$.
We still need an <idempotent ultrafilter> containing these sums; merely exhibiting an infinite set is insufficient. Put
$$
F_r=\operatorname{FS}(x_r,x_{r+1},\ldots),\qquad
K=\bigcap_{r\geq1}\overline{F_r},\qquad
\overline{F_r}=\{p:F_r\in p\}.
$$
The nonempty clopen sets $\overline{F_r}$ are nested, so compactness gives a nonempty compact $K$. This is the <tail finite-sums semigroup>. To verify closure under addition, take $p,q\in K$ and fix $r$. For any $s\in F_r$, choose a finite representation of $s$ and then an index $t$ beyond every index in that representation. Disjointness of supports gives $F_t\subseteq F_r-s$, so $F_r-s\in q$. Consequently
$$
\{s:F_r-s\in q\}\supseteq F_r\in p,
$$
and $F_r\in p+q$. This holds for every $r$, proving $p+q\in K$. The <Ellis–Numakura lemma> now supplies an <idempotent ultrafilter> in $K$. It contains $F_1$, a subset of the non-powers of $10$, so by upward closure it contains \b[the whole set of non-powers of $10$].
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