A criterion for finite abelian group to embed into a symmetric group

Yes, this is true (provided we add the hypothesis $i_1, \ldots, i_n \ge 1$), and not hard to prove by considering centralizers of products of cycles in symmetric groups.

According to http://www-history.mcs.st-andrews.ac.uk/Biographies/Povzner.html, the minimum degree of a permutation representation of an arbitrary abelian group was found in A. Ya. Povzner, Finding groups of permutations of least degree which are isomorphic to a given Abelian group, Kharkov Mat. Obshch., 14 (1937) 151–158. (I have not been able to obtain this paper to check.)

Theorem 4.1 in Kay Jin Lim, Specht modules with abelian vertices, J. Algebraic Combin. 35 (2012) 157–171 states your conjectured result in the special case when all the primes are the same with a reference to Povzner's paper.

Yes. Your notation $n$ is used for two different meanings, so let me denote by $m$ the number of summands. A homomorphism $G\to S_n$ correspond to an action of $G$ on a set with $n$ elements, which can be split into orbits $A_1,...,A_\ell$. Each orbit can can be provided with a structure of a group with a surjective homomophism $G\to A_i$. Then, the condition is precisely that the induced map $G\to \oplus_i A_i$ is injective. Note the total number of cyclic summands in the right hand side is at least $n$. If $A_i$ is not cyclic, using $ab\ge a+b$ for $a,b>1$, we see that it only decreases the sum of sizes if we split $A_i$ into two parts corresponding to a decomposition $A_i = B\oplus C$. So we may assume that all the $A_i$-s are cyclic of order a prime power. By comparing each primary component at a time, we may assume that all the $p_i$-s are a single prime $p$. The result now follows from the following standard lemma

Lemma: Let $\phi: \oplus_i \mathbb{Z}/p^{a_i} \to \oplus_j \mathbb{Z}/p^{b_j}$ be injective, with $a_1 \ge a_2 \ge ...\ge a_k$ and $b_1 \ge ...\ge b_\ell$. Then $a_i \le b_i$.

edit: As mensioned correctly by @Richard Lyons, the proof I gave for it was totally incorrect so I deleted it. Here's an hopefully less wrong proof, im really sorry for that.

Let $A\subseteq B$ be abelian finite $p$-groups. Write $B=\mathbb{Z}/p^k \oplus B'$ for $B'$ of exponent smaller than or equal $k$. Let $B\to \mathbb{Z}/p^k$ be the projection. If the composition $A\to B\to \mathbb{Z}/p^k$ is not surjective, then $A$ is contained in $p\mathbb{Z}/p^k \oplus B$ and we can proceed by induction onthe total number of elements of $B$. Otherwise $A$ contains an element of the form $(1,b)$ for $b\in B'$. Since the exponent of $B'$ is no more than $k$, there is a map $\mathbb{Z}/p^k\to B'$ mapping $1$ to $b$, and so by applying the automorphism $(x,c)\mapsto (x,c-xb)$ we may assume that $(1,0)$ is in $A$. It follows then that $A=\mathbb{Z}/p^k \oplus A\cap B'$ and we proceed by induction on the number of summands.