[...] the left-cancellative elements of P(S) are singletons. And the same reasoning can be now dualized to the case of right-cancellative elements, so we see that, at least on the level of groups, (P1) and (P2) are equivalent. (Hope to have added something to what you already knew!)

Assume $S$ is a grp (finite or infinite, abelian or not), and suppose to a contradiction that $X$ is left-cancellative for some $X\in P(S)$ with $|X|\ge 2$. Denote by $Y$ the subgrp of $S$ generated by $X$, pick $y\in Y$, and set $Z:=Y\setminus \{y\}$. Claim: $X+Y=X+Z$. Proof. It's enough to show that $X+y\subseteq X+Z$. For, fix $x_1\in X$ and, using $|X|\ge 2$, let $x_2\in X\setminus \{x_1\}$. Then, note that the equ. $x_1y=x_2z$ is solvable in $Z$, as (i) $x_1,x_2,y\in X$, (ii) $Y$ is a subgrp of $S$, and (iii) $x_2^{-1}x_1y\ne y$. This implies the claim, which gives in turn that [...]

@Michał. Right, I confused at least a couple of things with others. On a positive note, (P1) and (P2) can be proved to be equivalent if $S$ is a finite group (as far as I can say, you're interested also in the finite case, aren't you?), since then, however we choose $Y,Z\in P(S)$ with $|Y|+|Z| > |S|$, it holds $YZ=S$ (this is a folklore result of additive theory; e.g., it appears as Lemma 3.1.2 in the 3rd chapter of Geroldinger and Ruzsa's Combinatorial NT and Additive Group Theory), so that, if $X\in P(S)$ and $|X|\ge 2$, then $XS = XT = SX = TX = S$ for, say, $T:=S\setminus\{1\}$.

If $S$ is a left-cancellative finite sgrp (either commutative or not), then $SX=SS=S$ for every $X\in P(S)$, with the result that no element of $P(S)$ is left-cancellative, unless $S$ is empty or a singleton, in which case (P1) and (P2) are trivially equivalent. This means that, as far as the focus is on left-cancellative sgrps, you're left (excuse the unintended pun!) with the case where no element of $S$, other than the identity if an identity exists, has finite order.

Yes, sorry. I had assumed it was standard terminology. Given a cat $\sf C$, it's said that $\sf D$ is a full subcat of $\sf C$ if, well, $\sf D$ is a subcat of $\sf C$ and $\hom_{\sf D}(X,Y) = \hom_{\sf C}(X,Y)$ for all objects $X,Y \in {\rm Ob}(\sf D)$.

You should definitely replace the '$x$' appearing as an upper bound for the variable $\delta$ in the expression of the integral $I$ with '$t$'.

I'm finally convinced: there're first-countable topologies which are not semimetrizable (see the comments to mathoverflow.net/questions/163559 for details).

[...] About the origins of basic ideas in the area of asymmetric topology'' (in C. E. Aull and R. Lowen (eds.), Handbook of the History of General Topology, Vol. 3, Dordrecht: Kluwer (2001), 853-968), reports a letter by the same Fox where even a paracompact Hausdorff counterexample (to the $\gamma$-space conjecture) is provided.