Prove that the product over a sum is the sum over the cartesian products of the products. Prove this generalization of the distributive law,
$$ \prod_{a \in A} \sum_{b \in B_a} t_{a,b} = \sum_{c \in \prod_{a \in A} B_a}  \prod_{a \in A}t_{a,c_a} $$
where the product of sets $\prod_{a \in A} B_a$ is taken to be a cartesian product.
For example,
$$ A = \{ 1, 2 \} \wedge B_1 = \{ 1, 2 \} \wedge B_2 = \{ 1, 2 \} $$
$$ \prod_{a \in A} \sum_{b \in B_a} t_{a,b} $$
$$ = (t_{1,1}+t_{1,2}) (t_{2,1}+t_{2,2}) $$
$$ = \sum_{c \in \prod_{a \in A} B_a}  \prod_{a \in A}t_{a,c_a} $$
$$ = \sum_{c \in \{(m, n) | m \in B_1 \wedge n \in B_2\}}   t_{1,c_1}t_{2,c_2} $$
$$ = \sum_{c \in \{(1,1),(1,2),(2,1),(2,2)\}} t_{1,c_1}t_{2,c_2} $$
$$ = t_{1,1}t_{2,1}+t_{1,1}t_{2,2}+t_{1,2}t_{2,1}+t_{1,2}t_{2,2} $$

These definitions of the cartesian product are trivial, but useful.
Firstly, N-tuples are concatenated.
$$ (c_1, c_2)+(c_3, c_4) = (c_1, c_2, c_3, c_4) $$
R is a 1 dimensional cartesian product, in,
$$ R = \{(s)|s \in S\}$$
Use $\times$ for the cartesian product. If V and W are cartesian products,
$$ V\times W = \{v+w|v \in V  \wedge w \in W\}$$
If V and W are cartesian products,
$$ V\times W = \{v+w|v \in V  \wedge w \in W\}$$
If R, P, and Q are cartesian products,
$$  R\times P \cup R\times Q = R \times  (P \cup Q)$$
 A: For finite sums and products the statement is trivial. So I assume you meant generalizing to infinite sums and products.

You need absolute convergence. In the case of the Euler product for the Riemann zeta function, for $\sigma >1$, since $$\prod_p \frac{1}{1-p^{-\sigma}}=\prod_p (\sum_{k \ge 0} p^{-\sigma k})$$ is a product of series of non-negative terms, you are right it is enough to show that $0 < -\log(1-p^{-\sigma}) < p^{-\sigma}$ to obtain that $-\sum_p \log(1-p^{-\sigma})$ converges absolutely so that $\prod_p \frac{1}{1-p^{-\sigma}}$ converges absolutely, obtaining
$$\prod_p \frac{1}{1-p^{-\sigma}} = \prod_{i \ge 1} (\sum_{k \ge 0} p_i^{-\sigma k}) = \sum_{a \in \mathbb{N}^{\mathbb{N}_{\ge 1}}} \prod_{i\ge 1} p_i^{-\sigma a_i} = \sum_{n =1}^\infty n^{-\sigma}$$
where $\mathbb{N}^{\mathbb{N}_{\ge 1}}$ is the set of sequence of integers.

Now to prove rigorously such statements, it suffices to write infinite products and sums as what they are : limit of finite products and sums. Obtaining
 $$\prod_p \frac{1}{1-p^{-s}} \overset{def}= \lim_{m \to \infty} \prod_{i =1}^m \frac{1}{1-p_i^{-\sigma}}=\lim_{m \to \infty} \prod_{i =1}^m (\sum_{k \ge 0} p_i^{-sk}) \overset{(1)}=\lim_{m \to \infty} \sum_{a\in \mathbb{N}^m}\prod_{i=1}^m p_i^{-\sigma a_i}\\ = \lim_{m \to \infty} \sum_{\text{Lpf}(n) \le p_m} n^{-\sigma} \overset{(2)}= \lim_{N \to \infty}\sum_{n=1}^N n^{-\sigma}$$ 
where $(1)$ is justified because it is a finite product of series, and $\text{Lpf}(n) $ is the largest prime factor, and changing the order of summation in $(2)$ is justified by absolute convergence. 

Obviously the argument is the same in general, but the notations would be cumbersome, so I won't detail it.
A: For simplicity, let $A = \{1 .. n \}$. This reduces the complexity without loss of generality, as if we wanted to use an arbitrary set we could store the actual elements in an n-tuple and index them. So we can think of $1..n$ as representing the actual elements.
Use induction on $n$
For n = 1.
$$ \sum_{b \in B_1} t_{1,b} = \sum_{c \in B_1}  t_{1, c} $$
Which is true, proving the case.
Assume the formula is true for n, prove it for n+1.
$$ (\prod_{a \in \{1..n\}} \sum_{b \in B_a} t_{a,b})(\sum_{b \in B_{n+1}} t_{n+1,b}) = (\sum_{c \in \prod_{a \in \{1..n\}} B_a}  \prod_{a \in \{1..n\}}t_{a,c_a}) (\sum_{b \in B_{n+1}} t_{n+1,b}) $$
We need this result which we will prove below,
$$ (\sum_{c \in \prod_{a \in \{1..n\}} B_a}  \prod_{a \in \{1..n\}}t_{a,c_a}) \sum_{b \in X_{n+1}} t_{n+1,b} = \sum_{c \in (\prod_{a \in \{1..n\}} B_a) \times X_{n+1}}  \prod_{a \in \{1..n+1\}}t_{a,c_a}$$
Which gives,
$$ (\prod_{a \in \{1..n\}} \sum_{b \in B_a} t_{a,b})(\sum_{b \in B_{n+1}} t_{n+1,b}) = \sum_{c \in (\prod_{a \in \{1..n\}} B_a) \times B_{n+1}}  \prod_{a \in \{1..n+1\}}t_{a,c_a}$$
$$  = \sum_{c \in \prod_{a \in \{1..n+1\}} B_a}  \prod_{a \in \{1..n+1\}}t_{a,c_a}$$
Which completes the induction case on $n$, but assuming,
$$ (\sum_{c \in \prod_{a \in \{1..n\}} B_a}  \prod_{a \in \{1..n\}}t_{a,c_a}) \sum_{b \in X_{n+1}} t_{n+1,b} = \sum_{c \in (\prod_{a \in \{1..n\}} B_a) \times X_{n+1}}  \prod_{a \in \{1..n+1\}}t_{a,c_a}$$
To prove this, use induction on the size $X_{n+1}$.
If $X_{n+1}$ has a single element. 
$$ (\sum_{c \in \prod_{a \in \{1..n\}} B_a}  \prod_{a \in \{1..n\}}t_{a,c_a}) t_{n+1,k} = \sum_{c \in (\prod_{a \in \{1..n\}} B_a) \times \{k\}}  \prod_{a \in \{1..n+1\}}t_{a,c_a}$$
Which is true, proving the case for size 1.
Now assume it is true for $X_{n+1}$ and $Y_{n+1}$ we will prove it for $Z_{n+1}$. This corresponds to the induction step if the size of $Y_{n+1}$ is 1. So all we are doing is proving a more general case. We do this because $X_{n+1}$ and $Y_{n+1}$  are then symmetrical, and so the proof is easier to understand.
$$ Z_{n+1} = X_{n+1} \cup Y_{n+1} $$
Then
$$ \sum_{b \in Z_{n+1}} t_{n+1,b} = (\sum_{b \in X_{n+1}} t_{n+1,b} + \sum_{b \in Y_{n+1}} t_{n+1,b}) $$
Starting with the left hand side of the assumption,
$$ (\sum_{c \in \prod_{a \in \{1..n\}} B_a}  \prod_{a \in \{1..n\}}t_{a,c_a}) \sum_{b \in Z_{n+1}} t_{n+1,b} $$
Substituting for $Z_{n+1}$
$$ = (\sum_{c \in \prod_{a \in \{1..n\}} B_a}  \prod_{a \in \{1..n\}}t_{a,c_a}) (\sum_{b \in X_{n+1}} t_{n+1,b} + \sum_{b \in Y_{n+1}} t_{n+1,b}) $$
$$ = (\sum_{c \in \prod_{a \in \{1..n\}} B_a}  \prod_{a \in \{1..n\}}t_{a,c_a}) \sum_{b \in X_{n+1}} t_{n+1,b} + (\sum_{c \in \prod_{a \in \{1..n\}} B_a}  \prod_{a \in \{1..n\}}t_{a,c_a}) \sum_{b \in Y_{n+1}} t_{n+1,b}$$
Use the assumption for X and Y
$$= \sum_{c \in (\prod_{a \in \{1..n\}} B_a) \times X_{n+1}}  \prod_{a \in \{1..n+1\}}t_{a,c_a} + \sum_{c \in (\prod_{a \in \{1..n\}} B_a) \times Y_{n+1}}  \prod_{a \in \{1..n+1\}}t_{a,c_a}$$
Merging the sums,
$$= \sum_{c \in ((\prod_{a \in \{1..n\}} B_a) \times X_{n+1} \cup (\prod_{a \in \{1..n\}} B_a)\times Y_{n+1})}  \prod_{a \in \{1..n+1\}}t_{a,c_a}$$
The cartesian product is,
$$(\prod_{a \in \{1..n\}} B_a)\times X_{n+1} \cup (\prod_{a \in \{1..n\}} B_a)\times Y_{n+1}$$
Use $  R\times P \cup R\times Q = R \times  (P \cup Q)$
$$ = (\prod_{a \in \{1..n\}} B_a)\times (X_{n+1} \cup Y_{n+1})$$
Use $ X_{n+1} \cup Y_{n+1} = Z_{n+1} $
$$ = (\prod_{a \in \{1..n\}} B_a)\times Z_{n+1}$$
Substituting back in,
$$\sum_{c \in ((\prod_{a \in \{1..n\}} B_a)\times X_{n+1} \cup (\prod_{a \in \{1..n\}} B_a)\times Y_{n+1})}  \prod_{a \in \{1..n+1\}}t_{a,c_a} = \sum_{c \in (\prod_{a \in \{1..n\}} B_a)\times Z_{n+1}}  \prod_{a \in \{1..n+1\}}t_{a,c_a}$$
Which is the right hand side of the assumption. This completes the induction on size and completes the proof.
