Perhaps there is something relating odd ordered partitions with even ordered partitions?
There is indeed. Let's try to construct an involution $T_n$, mapping odd ordered partitions of $n$-element set to even and vice versa:
if partition has part $\{n\}$, move $n$ into previous part; otherwise move $n$ into new separate part.
Example: $(\{1,2\},\{\mathbf{5}\},\{3,4\})\leftrightarrow(\{1,2,\mathbf{5}\},\{3,4\})$.
This involution is not defined on partitions of the form $(\{n\},\ldots)$, but for these partitions one can use previous involution $T_{n-1}$ and so on.
Example: $(\{5\},\{4\},\{1,2\},\{\mathbf{3}\})\leftrightarrow(\{5\},\{4\},\{1,2,\mathbf{3}\})$.
In the end only partition without pair will be $(\{n\},\{n-1\},\ldots,\{1\})$. So our (recursively defined) involution gives a bijective proof of $\sum_{\text{k is even}}k!{n \brace k}=\sum_{\text{k is odd}}k!{n \brace k}\pm1$ (cf. 1, 2).
Upd. As for the second identity, the involution $T_n$ is already defined on all cyclically ordered partitions, so $\sum_{\text{k is even}}(k-1)!{n \brace k}=\sum_{\text{k is odd}}(k-1)!{n \brace k}$.
P.S. I can't resist adding that $k!{n \brace k}$ is the number of $(n-k)$-dimensional faces of an $n$-dimensional convex polytope, permutohedron (the convex hull of all vectors formed by permuting the coordinates of the vector $(0,1,2,\ldots,n)$). So $\sum(-1)^{n-k}k!{n \brace k}=1$ since it's the Euler characteristic of a convex polytope.
Best Answer
The definition $$ B_n = \sum_{k=0}^{n}{n\brace k} \tag{1}$$ imply that Bell number fulfill the following recurrence relation: $$ B_{n+1} = \sum_{k=0}^{n}\binom{n}{k}B_k \tag{2}$$ leading to their exponential generating function $$ f(x) = \sum_{n\geq 0}\frac{B_n}{n!}x^n = \exp\left(e^x-1\right).\tag{3} $$ Now, since $$ \exp\left(e^x\right) = 1+e^x+\frac{e^{2x}}{2!}+\frac{e^{3x}}{3!}+\ldots \tag{4}$$ the coefficient of $x^n$ in the RHS of $(4)$ is given by $$ \frac{1}{n!}+\frac{2^n}{n!2!}+\frac{3^n}{n!3!}+\ldots\tag{5}$$ and the claim (Dobinski's formula) $$ B_n = \frac{1}{e}\sum_{j\geq 0}\frac{j^n}{j!}\tag{6}$$ readily follows.