I'm interested in filling the detail of the claim I made above. I'm following Kochman's notation (page 14 for a def.).
Actually he never claims it, (he never spoke about ring spectra), but I think this is true, since for example it is true for $MO$, the Thom Spectrum for oriented bordism. In fact here on nLab, at the page of $(B,f)$-structure, it is claimed that the product must satisfy associatively and unitary, properties which should make $M\mathfrak{B}$ a ring spectrum.
So here is my problem: for my work I need to prove that the unit (which has to exists by axiom) $$i\colon \mathbb{S}\to M\mathfrak{B}$$ is the map induced by the inclusion of the base point $\ast \to B_n$, i.e. it becomes after applying the Thom space functor the inclusion of the fiber $i_n\colon S^n\to M\mathfrak{B}_n$.
What I tried so far
1) Since $\mathbb{S}$ is an initial object in RingSpec it’s equivalent to show that $i$ is a ring morphism but since the unit for $\mathbb{S}$ is the identity, showing that $i$ is a ring morphism involves showing that $i$ is the unit, so this method doesn’t provide any help.
2) Since the product is quite explicit on $BO$, I think I’m quite convinced that inclusion of fibre induced the unit in morphism $\mathbb{S} \to MO$. the problem is that I don’t know how to use naturality of the product to show that it must hold for $M\mathfrak{B}$ as well. By definition the product on $M\mathfrak{B}$ covers the one on $MO$ and so we have a morphism of ring spectra $Mf: M\mathfrak{B}\to MO$, therefore I can’t “push” the fiber inclusion (which is the identity on $MO$) to $M\mathfrak{B}$. (My aim was to use the fact that ring morphism send unit to unit, and $Mf$ should be an iso on fibres so restricted to inclusion is again the inclusion, but as I said I have maps $M\mathfrak{B}\to MO$ and not the other way round.
I think that what really is the problem is that Kochman never mentions the existence of a unit for such product, while on nLab it's claimed it exists. So I'm here to ask, what is the unit?
Kochman does mention existence of a unit, although in not a so transparent way. Have a look at the little diagram at page $14$, the second one. If you assume that $B_0=*$ (reasonable assumption in order to have fiber inclusion after thomifying) then you have your unit.
This is the diagram I'm mentioning, set $m=0$.
notice that $g_{n,n}$ is by def. the identity.