Specialize the Cycle Index Formula to get

$$\prod_{n=1}^\infty \exp \bigl( \frac{a_i}{i}z^i \bigr) = \sum_{n=0}^\infty \frac{z^n}{n!} \sum_{\pi \in \mathfrak{S}_n} a_1^{\mathrm{cyc}_1(\pi)} a_2^{\mathrm{cyc}_2(\pi)} \ldots, $$

by setting $a_i = t^2$ if $i$ is even and $a_i = t$ if $i$ is odd. We get

$$\sum_{n=1}^\infty \frac{z^n}{n!} \sum_{\pi \in \mathfrak{S}_n} t^{\mathcal{O}(\pi)+2\mathcal{E}(\pi)} = \frac{1}{(1-z^2)^{t^2/2}} \frac{(1+z)^{t/2}}{(1-z)^{t/2}}. $$

It is well known (see Corollary 7.21.4 in Stanley, *Enumerative Combinatorics II*) that
$$\prod_{\Box \in \lambda} \frac{d+c_\Box}{h_\Box} = s_\lambda(1^d)$$
is the number $|\mathrm{SSYT}_d(\lambda)|$ of semistandard Young tableaux of shape $\lambda$ with entries from $\{1,\ldots,d\}$, or equivalently, the dimension of the irreducible polynomial representation of $\mathrm{GL}_d(\mathbb{C})$ labelled by the partition $\lambda$.

Therefore it is equivalent to show that if $d \in \mathbb{N}$ then

$$ \frac{1}{(1-z^2)^{d^2/2}} \frac{(1+z)^{d/2}}{(1-z)^{d/2}} = \sum_{n=0}^\infty z^n \sum_{\lambda \in \mathrm{Par}(n) } |\mathrm{SSYT}_d(\lambda)| \tag{$\star$} $$

Now, as suggested by Gjerji Zaimi in a comment on my first answer, we use Littlewood's Identity. It states that

$$ \sum s_\lambda(x_1,\ldots,x_d) = \prod_{1 \le i < j \le d} \frac{1}{1-x_ix_j} \prod_{1 \le k \le d} \frac{1}{1-x_k} $$

where the sum is over all partitions $\lambda$. (For an online source, see (6) in this paper by Betea and Wheeler, and multiply by $\prod_{k=1}^d \frac{1}{1-x_kz} = \sum_{n=0}^\infty h_n(x_1,\ldots,x_d)z^n$ using Young's rule.) Specialize Littlewood's Identity by setting $x_i = z$ for all $i \in \{1,\ldots, d\}$ and sum over $n$ to get

$$ \sum_{n=0}^\infty \sum_{\lambda \in \mathrm{Par}(n)} s_\lambda(1^d) z^n = \frac{1}{(1-z^2)^{(d^2-d)/2}(1-z)^d}. $$

The right-hand side agrees with the left-hand side of ($\star$), so we are done.