$\newcommand{\cat}{\mathrm} \newcommand{\St}{\cat{Stone}^\cat{fr}} \newcommand{\Cat}{\cat{Cat}} \newcommand{\Cart}{\cat{Cart}} \newcommand{\Fun}{\cat{Fun}} \newcommand{\Mon}{\cat{Mon}} \newcommand{\Set}{\cat{Set}} \newcommand{\Po}{\cat{Poset}}$ EDIT : I misread the definition of ultracategory fibration, apparently only certain locally cartesian edges are closed under composition. After a thought about it, what I said for monoids remains true, but not for posets. I just need to modify the arguments.
I would argue that a monoid is not the same as a category with one object, namely a monoid is the same as a pointed category with one object. This is not really relevant if you look at the category of categories as a $1$-category (because then $\hom_\Cat(BM, BM')$ is indeed isomorphic to $\hom_\Mon(M,M')$), but it does if you more naturally view it as a $(2,1)$-category ($\Fun(BM,BM')$ has nontrivial morphisms !)
In particular, I'll interpret your question as :
What is an ultracategory $\mathcal M$ with a morphism from the terminal ultracategory "$*$" such that for all $I$, $*_{\beta I}\to \mathcal M_{\beta I}$ is essentially surjective ?
Of course, because $*_{\beta I}\simeq \prod_I*$ in a way compatible with $\mathcal M_{\beta I}\simeq \prod_I\mathcal M_*$, this is equivalent to $*\to \mathcal M_*$ being essentially surjective.
The first question is : What is a pointed ultracategory ?
In my original post, I made a mistake because I had misread the definition of ultrcategory. In particular, a morphism $\St\to \mathcal M$ is more than just a point in $\mathcal M_*$ : it's a point with a certain property. Namely, (see definition 1 in Lurie's notes) we ask that the comparison morphisms for base-change functors associated to all compositions $\beta I\to \beta J\to \beta K$ be isomorphisms when applied to this point (while in the definition of an ultracategory, one only requires this if $\beta I \to \beta J$ comes from a map $I\to J$).
But, in particular, if this point is the only point of $\mathcal M_*$ (and thus of $\mathcal M_{\beta I}$ for all $I$), this amounts to asking that cartesian morphisms actually be closed under composition ! In other words,
An ultramonoid is an ultracategory which is a cartesian fibration $\mathcal M\to \St$ with a section $\St\to \mathcal M$ sending cartesian edges to cartesian edges, such that on fibers over $*$, the map $*\to \mathcal M_*$ is essentially surjective.
Now that this is clarified, the rest of my answer essentially goes through without changes. But note that there was initially a mistake, and also note that this will not work for posets !
The second thing to observe is that there is a functor $\Mon\to\Cat_*$ which implements the equivalence of $(2,1)$-categories I've mentioned above.
This functor is fully faithful (in fact it has a right adjoint - given by $(C,x)\mapsto \cat{End}_C(x)$, this will be relevant later - and the unit map $M\to \cat{End}_{BM}(\bullet)$ is an isomorphism), and therefore the functor $\Fun((\St)^\cat{op}, \Mon)\to \Fun((\St)^\cat{op},\Cat)$ is also fully faithful.
By what was clarified above, ultramonoids can therefore be viewed as a full subcategory of $\Fun((\St)^\cat{op},\Cat_*)$, via the Grothendieck construction (while this is not the case for general ultracategories ! for them it would be certain pseudo-functors). Precisely, it is the full subcategory spanned by functors $(\St)^\cat{op}\to\Cat_*$ that send the diagrams $(\{i\}\to \beta I)_{i\in I}$ to product diagrams. Piecing things together, using the fact that $\Mon\to\Cat_*$ and $\Cat_*\to\Cat$ both reflect and preserve products, we find :
The category of ultramonoids is equivalent to the full subcategory of $\Fun((\St)^\cat{op},\Mon)$ spanned by those functors that send the diagrams $(\{i\}\to \beta I)_{i\in I}$ to product diagrams.
But we can be more concrete. Indeed, $\Fun((\St)^\cat{op},\Mon)\simeq \Mon(\Fun((\St)^\cat{op},\Set))$, and $\Mon \to \Set$ preserves and reflects products, so that :
The category of ultramonoids is equivalent to the category of monoids in the category of ultrasets.
(note that for ultrasets the subtlety disappears: the fibrations must be cartesian, and not only locally cartesian)
Ultrasets are compact Hausdorff topological spaces, so finally:
Corollary : The category of ultramonoids is equivalent to the category of monoids in compact Hausdorff topological spaces.
Now, if $\mathcal M$ is an ultracategory with a point $x : *\to \mathcal M_*$, there is, in general, no way to make an ultramonoid out of it because of this cartesian vs locally cartesian business. However, if it is a "nice" point, i.e. if it is a point for which the comparison morphisms for general compositions $\beta I\to \beta J\to \beta K$ are isomorphisms, then we can view it as a morphism $\St\to \mathcal M$ of ultracategory fibrations. In particular we can take the full subcategory $\mathcal M_x$ of $\mathcal M$ spanned by the image of $\St$, and this should still be an ultracategory fibration. But now this is a pointed ultracategory fibration with an essentially surjective point, so it is an ultramonoid, as explained above.
We therefore find tha a sufficiently nice object $x$ (in more details: a point $\St\to \mathcal M$) in an ultracategory $\mathcal M$ has an ultramonoid of endomorphisms, $\cat{End}_\mathcal M(x)$, i.e. a compact Hausdorff topological space of endomorphisms.
(For posets, I don't know a satisfying answer. I'm not sure you will get a more satisfying answer than to the question "what is an ultracategory ?", but I would be interested in seeing one)