Not an answer, but too long for a comment. This addresses a previous comment and points out the smallest case where the requested bijection is not obvious (for the even case).
Through $n = 8$, the partition sets are in bijection via a combination of transposition and the identity map, so there is a direct correspondence for the tableaux.
$Pe_{10} = \{10, 82, 64, 622, 442\}$ and $Qe_{10} = \{55, 4411, 331^4, 221^6, 1^{10}\}$. The partitions that don't match up are 64 & 442 versus 55 & 4411.
Using the hook length formula, there are 90 standard Young tableaux of shape (6,4) and 252 of shape (4,4,2), so these contribute 342 to $TPe_{10}$. (Sorry for the change in partition notation, but "90 of shape 64" looked strange.) There are 42 tableaux of shape (5,5) and 300 of shape (4,4,1,1), also contributing 342 to $TQe_{10}$. So the conjecture does hold for $n=10$.
This example shows that the requested bijection cannot preserve the tableaux shape/underlying partition.
The next case is even more interesting because the tableaux counts match even though there are not the same number of partitions in the sets.
The unmatched partitions are $(8,4),(6,4,2),(4,4,4) \in Pe_{12}$ and $(5,5,1,1),(4,4,1,1,1,1) \in Qe_{12}$. And the sums work: Writing $\textrm{SYT}(\lambda)$ for the number of standard Young tableaux with shape $\lambda$, we have
$$\textrm{SYT}((8,4))+\textrm{SYT}((6,4,2))+\textrm{SYT}((4,4,4))=275+2673+462=3410,$$
$$\textrm{SYT}((5,5,1,1))+\textrm{SYT}((4,4,1,1,1,1))=1485+1925=3410.$$
The next two cases work but don't offer anything new. Here are a few general comments.
From a formula for the number of partitions of $n$ with up to three parts, $|Pe_{2n}| = \lfloor \frac{(n+3)^2}{12} \rceil$ where $\lfloor \cdot \rceil$ denotes nearest integer.
For $n$ odd, 3 partitions match up between the two sets: $(2n)$, $(2n-2,2)$, and $(2n-4,2,2)$ in $Pe_{2n}$ with $1^{2n}$, $221^{2n-4}$, and $331^{2n-6}$ in $Qe_{2n}$ (matching by transposition).
For $n$ even, 4 partitions match because $(n,n)$ is in both sets.
Continuing along this line of reasoning would require formulas for $\textrm{SYT}(\lambda)$ for partitions from each of the sets.
Following up on the last bullet, in the 1996 JCTA note "Polygon Dissections and Standard Young Tableaux," Stanley credits O'Hara and Zelevinsky with a formula for $\textrm{SYT}(\lambda)$ for $\lambda = kk1^m$. Rewritten in terms of $k$ and $m$, it is
$$\frac{1}{2k+m+1} \binom{2k+m+1}{k} \binom{m+k-1}{k-1}.$$ Perhaps Stanley's bijection between such tableaux and certain sets of nonintersecting diagonals in a polygon discussed in the note could help.
It's not hard to work out that
$$\textrm{SYT}((a,b,c))=\frac{(a+b+c)!(a-b+1)(b-c+1)(a-c+2)}{(a+2)!(b+1)!c!}$$
(without regard to the parity of $a,b,c$).
Best Answer
The sum $\sum_\mu \chi^\lambda_\mu$ over partitions $\mu$ of $n$ is the multiplicity of the irreducible $\chi^\lambda$ in the character afforded by $\mathfrak S_n$ acting on itself by conjugation. If $\psi$ is the character for conjugation, then $\psi(g)$ is the size of the centralizer of $g$, so $$\langle \chi^\lambda,\psi\rangle =\sum_{\mu\vdash n} \frac{\chi^\lambda_\mu z_\mu}{z_\mu},$$ where $z_\mu$ is the size of the centralizer of any permutation with cycle type $\mu$. So your sum is counting how many irreducibles the conjugation representation breaks up into.
I should mention that the individual row sums $\sum_{\mu\vdash n} \chi^\lambda_\mu$ are not completely understood. It is already non-trivial to prove that they are greater than $0$ (this should be in "An explicit model for the complex representations of $S_n$" by Inglis, Richardson and Saxl.) This is unlike the situation for the group acting on itself by left multiplication. However, when you take your double sum, then there is a good combinatorial description. In fact, summing down any column we know: $$\sum_\lambda \chi^\lambda_\mu = \# \{g\in \mathfrak S_n : g^2=h\}$$ for any fixed $h\in\mathfrak S_n$ with cycle type $\mu$. This works because each irreducible character of $\mathfrak S_n$ is the character of a real representation. In general, there is the work of Frobenius--Schur. (This is what Lucia was already quickly and correctly observing. Indeed, the sum of degrees will dominate.)
For your second question, it comes from the paper linked to by Lucia in your previous post. Miller proved the statement you are observing (see Theorem 2 and its proof, particularly the last few lines). The number of even entries plus the number of odd entries (denoted $E_n$ and $O_n$) is congruent to the number of partitions $p(n)$, and $E_n$ is always even. So $O_n\equiv p(n)$ mod 2.