Proof that the number of 1's in $P(n)$ equals the number of distinct magnitudes in $P(n)$
This could be solved by considering generating functions. Call the first sum $\mathcal A_n$ and the second sum $\mathcal B_n$.
Thus first we have: $$\begin{align*} \sum_{n\in\mathbb{N}} \mathcal A_nx^n&= \underbrace{\left(\sum_{k\ge0}kx^k\right)}_{\text{Counting # of 1's}}\underbrace{\left(\prod_{r\ge2}\frac1{1-x^r}\right)}_{\text{The rest}}\\ &= \frac x {\left ( 1-x \right )^2}\,\left(\prod_{r\ge2}\frac1{1-x^r}\right)\\ &=\frac x {1-x}\,\left(\prod_{r\ge1}\frac1{1-x^r}\right) \end{align*}$$ Then: $$\begin{align*} \sum_{n\in\mathbb{N}} \mathcal B_nx^n&= \frac{\partial }{\partial k}\left.\left ( \prod_{r\ge1}\left ( 1+kx^r+kx^{2r}+kx^{3r}+\cdots \right ) \right )\right|_{k=1}\\ &= \frac{\partial }{\partial k}\left.\left ( \prod_{r\ge1}\left ( \frac k {1-x^r}-k+1 \right ) \right )\right|_{k=1}\\ &=\left.\left ( \prod_{r\ge1} \frac k {1-x^r}-k+1 \right )\underbrace{\left ( \sum_{r\ge1} \frac{\frac 1 {1-x^r}-1}{\frac k {1-x^r}-k+1} \right )}_{\text{Product rule term}}\right|_{k=1}\\ &=\left ( \prod_{r\ge1} \frac 1 {1-x^r} \right )\left ( \sum_{r\ge1} x^r \right )\\ &=\frac x {1-x}\,\left(\prod_{r\ge1}\frac1{1-x^r}\right) \end{align*}$$ Therefore $\mathcal A_n=\mathcal B_n$ as desired.
As a side note, this also suggests that both are equal to $\sum_{h=0}^{n-1}P(h)$, so maybe there is a more elementary approach.
Update:
Turns out there is! For the following I'll write $\mathcal P_n=\sum_{h=0}^{n-1}P(h)$.
$\text{1. }\mathcal A_n=\mathcal P_n\text{:}$
For every partition of $n$ containing $k$ many $1$'s, we map it to $k$ other partitions as follows, for example: $$\color{red}{1+1+1}+3+4~\longrightarrow~\begin{cases} \color{blue}{1+1}+3+4 \\ \color{blue}{1}+3+4 \\ 3+4 \end{cases}$$ Thus the reverse mapping is by appending "$+1$"s to any partition of some $0\le h\le n-1$ until the whole sums evaluates to $n$.
Therefore the sum of all appearance of $1$'s is equal to the count of all partitions of any $0\le h\le n-1$, which shows that $\mathcal A_n=\mathcal P_n$ as desired.
$\text{2. }\mathcal B_n=\mathcal P_n\text{:}$
We can obviously transform $\mathcal B_n$ to the sum $\sum_{k=1}^{n}P\left(n\mid\exists k\right)$ where $P\left(n\mid\exists k\right)$ counts the number of partitions of $n$ using at least one $k$.
Note however that $P\left(n\mid\exists k\right)$ is just $P\left(n-k\right)$, thus $$\mathcal B_n=\sum_{k=1}^{n}P\left(n-k\right)=\sum_{h=0}^{n-1}P(h)=\mathcal P_n$$ as desired.