Is $(1+2+3+…)=(1+2+2^2+2^3+…)(1+3+3^2+…)(1+5+5^2+…)…$? Are these equal?
$$(1+2+3+…)=(1+2+2^2+…)(1+3+3^2+…)(1+5+5^2+…)…$$
Where the RHS has a series for each prime. Looks like they are the same series by the fundamental theorem of arithmetic.
Every number on the LHS is a product of prime factors which can be obtained from the right hand side by choosing the appropriate numbers from each contributing prime and $1$ from the rest and conversely.
By a similar argument, it looks like:
$$(1+2+3+\dots)=(1+2+3+5+\dots)(1+2+3+5+\dots)(1+2+3+5+\dots)\dots$$

Maybe you can try proving
$$
(1^s+2^s+3^s+\dots)=(1^s+2^s+2^{2s}+\dots)(1^s+3^s+3^{2s}+\dots)(1^s+5^s+5^{2s}+\dots)\dots
$$
for $s$ such that the series converge.
The left side is $\zeta(-s)$, the factors on the right side are geometric series.  I think your result will be a well-known formula.  But there are analytic continuations for all of these to other values of $s$ where the series diverge.  Put $s=1$ to get a conclusion, $\zeta(-1) = \dots$.
