As shorthand, let $\sum_\pi(n)$ denote the sum of prime indices of $n$ as described above.
For any $k$, we know $\pi(p_k)=\sum_\pi(p_k)=k$.
From Bertrand's postulate, we know that $\pi(2^k) \geq \sum_\pi(2^k)=k$.
Any intermediate $n$ such that $\sum_\pi(n)=k$ will fall in the range $(p_k,2^k)$ and thus have $\pi(n) \geq k$. This can be shown explicitly by starting with $2^k$ and repeatedly dividing by $2$ while increasing another factor to its next larger prime. Bertrand ensures that the next larger prime will always be less than the factor of $2$ we lost, and so the overall product must also decrease.
e.g. $(1,1,1,1,1)\rightarrow (2,1,1,1)\rightarrow (2,2,1)\rightarrow (3,2)$, using integer partition tuples to represent the products $2^5=32$, $3\cdot 2^3=24$, $3^2\cdot 2=18$, $5\cdot 3=15$, and the minimum we could have gone to here is $(5)=p_5=11$.
Thus we see that any $n$ where $\sum_\pi(n)=k$ must be in the range $p_k \leq n \leq 2^k$, and since $\pi$ is a monotonically increasing function, we know that any such intermediate $n$ will give $\pi(p_k) \leq \pi(n) \leq \pi(2^k)$. And again, since $\sum_\pi(p_k)=k$, any $n>p_k$ will have $\pi(n)\geq k$.
This should suffice to prove $\pi(n) \geq \sum_\pi(n)$ for all $n\in\mathbb N$, or equivalently, that $\pi(ab)\geq \pi(a)+\pi(b)$.
To get the bounded inequality given in the problem, I think it's a matter of how many special cases you want to address; the larger the lower bound for $n$, the larger the constant you can reliably add to the right side of this, i.e. for $n\geq 12$, you can use $\pi(n) \geq \sum_\pi(n)+1$. In other words, for any $c$, there's some $N$ such that for all $n \geq N$, you get $\pi(n) \geq \sum_\pi(n)+c$.