= Solution
Call $A\subseteq\mathbb N$ an <IP set> when it contains an infinite <finite-sums set> $\operatorname{FS}(x_1,x_2,\ldots)$ with all $x_i>0$. We first need the relative consequence of the <Hindman theorem>: \b[a finite union can be an <IP set> only if one of its constituents is an <IP set>]. This does not follow by simply applying the theorem to a coloring of all integers, because its <monochromatic> <finite-sums set> might lie outside the given set.
Here is a reduction that proves the <partition regularity of IP sets>. Suppose $\operatorname{FS}(x_i)\subseteq A$ and $A$ is finitely colored. Index the generators by $i\geq0$. For $n>0$, let $S(n)$ be the set of positions of the nonzero digits in its <binary expansion>, and color $n$ by the color of $\sum_{i\in S(n)}x_i$. By the <Hindman theorem>, there are $w_1,w_2,\ldots>0$ with $\operatorname{FS}(w_i)$ <monochromatic> in this new <finite coloring>.
We can take disjoint, successively later finite blocks of the $w_i$ whose sums $v_1,v_2,\ldots$ have separated <binary expansion> supports. Start with $v_1=w_1$. Having chosen finitely many blocks, take $K$ larger than every binary position used in their sums. On a fresh tail of the $w_i$, consider $2^K+1$ partial sums, including the zero partial sum. Two have the same <residue class> modulo $2^K$, by the <pigeonhole principle>. Their positive difference is a consecutive block sum $v_{j+1}$ divisible by $2^K$. Discard all indices up to that block's end and repeat. Thus every bit of $v_{j+1}$ is above every bit in the preceding block sums.
There are no carries in a sum of distinct $v_j$, so
$$
S\left(\sum_{j\in J}v_j\right)=\bigsqcup_{j\in J}S(v_j)\qquad(\varnothing\ne J\text{ finite}).
$$
Also $\operatorname{FS}(v_j)\subseteq\operatorname{FS}(w_i)$, since the underlying blocks are disjoint. Consequently
$$
y_j:=\sum_{i\in S(v_j)}x_i
$$
satisfies $\operatorname{FS}(y_j)\subseteq\operatorname{FS}(x_i)\subseteq A$, with all these sums the same original color. This proves the relative assertion. In particular, if $A_1\cup\cdots\cup A_r$ contains an infinite <finite-sums set>, color its points by the first constituent containing them and apply this argument; one $A_j$ is an <IP set>.
Let $\mathcal I$ be the family of sets that are not <IP sets>. It is closed under subsets and, by the preceding argument, under finite unions. It contains every <finite set>, since positive successive partial sums of an infinite sequence are unbounded. It does not contain $\mathbb N$. Therefore
$$
\mathcal F=\{A\subseteq\mathbb N:A^c\in\mathcal I\}
$$
is a proper <filter on a set>: it contains $\mathbb N$, excludes the empty set, is upward closed, and is closed under finite intersections. Its members are exactly the <IP-star sets>, the sets meeting every <IP set>. Indeed, failure to meet an <IP set> is equivalent to having an IP complement. This identifies the sets that must belong to the required <ultrafilter>.
Apply the <ultrafilter lemma> to extend the <IP-star filter> $\mathcal F$ to an <ultrafilter> $\mathcal U$. For completeness, <Zorn lemma> applies to its proper filter extensions ordered by inclusion, because the union of a chain is still a proper <filter on a set>. A maximal extension decides every set $A$: if $A$ cannot be adjoined while keeping the filter proper, some filter member is disjoint from $A$, forcing $A^c$ into the filter. This is the defining <ultrafilter> property.
If $A\in\mathcal U$ were not an <IP set>, then $A^c\in\mathcal F\subseteq\mathcal U$, contradicting properness. Hence \b[every member of $\mathcal U$ contains an infinite <finite-sums set>]. Conversely, any <ultrafilter with finite-sums members> must contain every <IP-star set>, since it cannot contain such a set's non-IP complement. Thus this construction captures precisely the forced filter, without making an unnecessary claim about idempotence.
To prove nonuniqueness, put
$$
E=\bigcup_{j\geq0}\bigl([2^{2j},2^{2j+1})\cap\mathbb N\bigr).
$$
For a nonempty finite set of exponents with largest exponent $J$, the associated sum of distinct powers of four satisfies
$$
4^J\leq\sum_{j\in F}4^j\leq\frac{4^{J+1}-1}{3}<2\cdot4^J.
$$
Thus $\operatorname{FS}(4^j:j\geq0)\subseteq E$. Doubling the inequality gives
$$
2\cdot4^J\leq\sum_{j\in F}2\cdot4^j<4\cdot4^J,
$$
so $\operatorname{FS}(2\cdot4^j:j\geq0)\subseteq E^c$. These <alternating dyadic intervals contain disjoint IP sets>.
Every member of $\mathcal F$ meets $E$ and also $E^c$, because it meets every <IP set>. A finite intersection of members of $\mathcal F$ is again a member, so both $\mathcal F\cup\{E\}$ and $\mathcal F\cup\{E^c\}$ have the <finite intersection property>. Extend the proper filters they generate to <ultrafilters> $\mathcal U_0$ and $\mathcal U_1$. Each extends $\mathcal F$, so the previous argument makes each an <ultrafilter with finite-sums members>. But $E\in\mathcal U_0$ and $E^c\in\mathcal U_1$, so
$$
\boxed{\mathcal U_0\ne\mathcal U_1.}
$$
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