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@AlexanderChervov I see, what you are saying is that the definition of the Segre product uses the braiding. That is indeed correct, and one has to look with some care at what is going on. It is not covered in [PP], but I do believe that the result is true, though it is probably faster to prove it yourself than to try to look for a reference.
@AlexanderChervov 1) The Manin black product by design can be used to compute the internal cohom for quadratic algebras, so yes, everything works the same for two different algebras. 2) Your superalgebras are still associative algebras with quadratic relations, and the symmetric monoidal structure does not interfere with the Koszul property, so everything applies literally.
Just the fact that you think that some new definition is better than the existing one is not quite a sufficient reason to arbitrarily redefine existing notions, no? I think the question was "what is a cogroup", rather than "how would you like to redefine the notion of a cogroup"...
Section 3 of Markl's paper begins with "As we mentioned in the introduction, by a strongly homotopy P-algebra we mean an algebra over a minimal, or at least cofibrant, resolution of the operad P." - I would assume that the constrains imposed on the operad are motivated mainly by thinking about existence of minimal models. If you are have a minimal model of Sullivan type (which seems to be more or less what Markl calls elementally cofibrant), all the usual arguments go through mutatis mutandis, I believe.
The paper of Markl suggests to look at the minimal model. If $P$ has a nontrivial differential, or if $P(1)$ is not the ground field, the minimal model may not exist. So in some sense the answer to your question is "no, for trivial reasons". Would you like to be more precise perhaps?
@AlexanderBraverman In arxiv.org/abs/2112.12436 Nicolas Perrin and Maxim Smirnov state, in Lemma 4.6, part 3, a simple formula for the degree of the Schubert class corresponding to the root $\alpha$, leading to a simple formula for the Poincaré polynomial. (They refer to the article cited in my answer where this formula is not proved, but I think it should be easy to prove it directly - I just do not have time for this at the moment.)
Moreover, the answer obtained by @FredHucht admits an immediate proof. Subtract from the first line $x_1$ times the second, develop along the first row, and do the same with the remaining matrix. Every time you earn $1-x_1x_2\cdots x_n$ as a factor.
