Given any fraction $\frac{s}{t}=\frac{s}{\Pi_pp^{i_p}}$ with $s,t$ relatively prime, I would like to know if it is possible to write $\frac{s}{\Pi_pp^{i_p}}=\sum_p\frac{s_p}{p^{i_p}}$ for some unique integers $s_p$.
It seems like this is a simple algebraic fact, but I'm not totally sure how to prove it, especially the uniqueness part. The furthest that I've gotten is that for all nonzero $i_p$ we must have $s_p$ nonzero, because otherwise $s$ would be divisible by that $p$, contradicting that $t$ and $s$ are relatively prime.
I guess an equivalent question would be to ask for solutions to some finite equation $\sum_i a_ix_i=a$ where $\text{gcd}(a,\{a_i\})=1$.
No, this is not true. We can consider the simple case where $t=pq$ where $p,q$ are distinct primes, and $s=1$. Here, we choose $s_p, s_q$ to satisfy $ps_q + qs_p=1$, and such $s_p, s_q$ exist per Bezout's identity. Then observe that
$$\frac{s_p}{p} + \frac{s_q}{q} = \frac{qs_p + ps_q}{pq} = \frac{1}{pq} = \frac{s}{t}$$
In general, numerous $s_q, s_p$ can be found which satisfy $ps_q + qs_p=1$. For instance, take $s_p'=s_p-pk$, $s_q' = s_q+qk$ where $k$ is an arbitrary integer, so that $ps_q' + qs_p' = p(s_q+qk) + q(s_p-pk)=ps_q + qs_p = 1$.