Cycle generating function of permutations with only odd cycles
Conjecture 2 follows from setting $t=2k$ in the formula $$ \sum_{n\geq 0}\mathcal{C}_{\mathrm{ODD}}(n,t)\cdot\frac{z^n}{n!} =\left(\frac{\sqrt{1-z^2}}{1-z}\right)^t. $$ It then follows easily that $$ \sum_{j\geq 0}P_k(j)x^j = \frac{(2k-1)!!\,x(1+x)^{k-1}}{(1-x)^{k+1}}.\ \qquad (1) $$ By Theorem 3.2 of http://math.mit.edu/~rstan/papers/cycles.pdf we see that all the zeros of $P_k(x)$ are purely imaginary, which implies that $P_k(x)$ has nonnegative coefficients and is either even or odd, depending on the parity of $k$. Clearly also from (1) $P_k(0)=0$. I don't see immediately from (1) why $P_k(x)$ has integer coefficients.
Let me also remark that the polynomial $(P_k(x)+P_k(x+1))/(2k-1)!!$ is the Ehrhart polynomial of the standard $k$-dimensional cross-polytope. See Exercise 4.61 of Enumerative Combinatorics, vol. 1, second edition.
I'll start by addressing conjecture 2. By summing your generating fun over all values of $k$ we obtain $$F(z,w)=\sum_{n\geq 0}\sum_{k\geq 0}C_{\text{ODD}}(n,2k)\frac{z^n}{n!}w^k=\sum_{k\geq 0}w^k\left(\frac{1+z}{1-z}\right)^k=\frac{1-z}{1-w-z-wz}$$ From here we see that $$\sum_{n\geq 1}\sum_{k\geq 1}\frac{P_k(n)}{(2k-1)!!}w^k \frac{z^{n-1}}{(n-1)!}=\frac{1}{2}\frac{d}{dz}\int\left(F(z,w)-1\right)dw$$ $$=\int \frac{w}{(1-w)^2}\frac{1}{\left(1-z(\frac{1+w}{1-w})\right)^2}dw$$ By extracting the coefficients of $z^{n-1}/(n-1)!$ on both sides we have $$\sum_{k\geq 0}\frac{P_k(n)}{(2k-1)!!}w^k=\int \frac{nw}{(1-w)^2}\left(\frac{1+w}{1-w}\right)^{n-1} dw=\frac{1}{2}\left(\frac{1+w}{1-w}\right)^n+\text{constant}$$ which is what we wanted.
I just realized that we can also answer conjecture 1 by making use of this identity. Start with the expansion $$\left(\frac{1+w}{1-w}\right)^n=\left(1+\frac{2w}{1-w}\right)^n=1+\sum_{r\geq 1} \binom{n}{r}\left(\frac{2w}{1-w}\right)^r$$ $$=1+\sum_{k\geq 1}w^k\sum_{r\geq 1}2^r\binom{n}{r}\binom{k-1}{r-1}$$ which gives us an explicit formula for $P_k$ $$P_k(n)=\sum_{r\geq 1}2^{r-1}(2k-1)!!\binom{n}{r}\binom{k-1}{r-1}$$ Which immediately tells us that $P_k(n)$, as a linear combination of $\binom{n}{r}$ for $1\le r\le k$, is a polynomial of degree $k$ with no constant term. Combined with the fact that $\left(\frac{1+w}{1-w}\right)^n=\left(\frac{1-w}{1+w}\right)^{-n}$ we have $P_k(n)=(-1)^kP_k(-n)$ which tells us that $P_k$ has the same parity as $k$. Now it remains to establish integrality of the coefficients. The explicit formula for $P_k$ can be rearranged as $$P_k(n)=\sum_{r\geq 1}\binom{k+r-1}{r,r-1,k-r}\frac{(2k-1)!}{2^{k-r}(k+r-1)!}(n)_r$$ and from here it is clear that the coefficients of $P_k$ have nonnegative $p$-adic valuation for any odd prime $p$. It remains to show the following lemma
Lemma: For any $k\geq r\geq 1$ we have $$\nu_2\left(\frac{(2k-1)!}{r!(r-1)!(k-r)!}\right)\geq k-r.$$ Proof We make use of the fact that $\nu_2\left(\frac{s!}{\lfloor\frac{s}{2}\rfloor!}\right)=\lfloor\frac{s}{2}\rfloor$. Our expression can be written as $$\nu_2\left(\frac{(2k-1)!}{r!(r-1)!(k-r)!}\right)=\nu_2 \left(\frac{(2k-1)!}{(k-1)!}\frac{\lfloor\frac{r}{2}\rfloor!\lfloor\frac{r-1}{2}\rfloor!}{r!(r-1)!}\right)+\nu_2\left(\binom{k-1}{\lfloor\frac{r}{2}\rfloor,\lfloor\frac{r-1}{2}\rfloor,k-r}\right)$$ the first term is equal to $k-1-\lfloor\frac{r}{2}\rfloor-\lfloor\frac{r-1}{2}\rfloor=k-r$ and the second term is clearly nonnegative. This completes the proof of integrality.