Note that we can write $\displaystyle A = 1 - \frac1{3^n} + \frac1{5^n} - \frac1{7^n} + \frac1{9^n} - \frac1{11^n} + \frac1{13^n} - \cdots$ as $A = \left(1 - \frac1{3^n} + \left(\frac1{3^n} \right)^2 - \cdots \right) \left(1 + \frac1{5^n} + \left(\frac1{5^n} \right)^2 + \cdots \right)\left(1 - \frac1{7^n} + \left(\frac1{7^n} \right)^2 - \cdots \right) \times$ $ \left(1 - \frac1{11^n} + \left(\frac1{11^n} \right)^2 - \cdots \right) \left(1 + \frac1{13^n} + \left(\frac1{13^n} \right)^2 + \cdots \right)\cdots$ Since every odd number of the form $4k+3$ must have odd number of prime factors of the form $4k+3$ counted with multiplicity and every every odd number of the form $4k+1$ must have even number of prime factors of the form $4k+3$ counted with multiplicity.
EDIT
We will now prove what you have stated for $n>1$.
If we let $ F(n) = 1 - \frac1{3^n} + \frac1{5^n} - \frac1{7^n} + \frac1{9^n} - \frac1{11^n} + \frac1{13^n} - \frac1{15^n} + \frac1{17^n}\cdots,$ then $ \frac{F(n)}{3^n} = \frac1{3^n} - \frac1{9^n} + \frac1{15^n} - \frac1{21^n} + \frac1{27^n} - \frac1{33^n} + \frac1{39^n} - \frac1{45^n} + \frac1{51^n}\cdots$
Hence, $F_3(n) = F(n) \left( 1 + \frac1{3^n} \right) = F(n) + \frac{F(n)}{3^n} = 1 + \frac1{5^n} - \frac1{7^n} - \frac1{11^n} + \frac1{13^n} + \frac1{17^n} - \frac1{19^n} - \cdots$
By doing the above, we have now removed all the multiples of $3$. Note that while adding we are allowed to rearrange and add since we have assumed $n>1$ and we know that for $n>1$, $F(n)$ converges absolutely. Repeat the process for $F_3(n)$ using the rest of the primes by adding or subtracting depending on whether the prime is $\pm 1 \bmod 4$.
For instance, $\frac{F_3(n)}{5^n} = \frac1{5^n} + \frac1{{25}^n} - \frac1{{35}^n} - \frac1{{55}^n} + \frac1{{65}^n} + \frac1{{85}^n} - \frac1{{95}^n} - \cdots$
Subtract the above from $F_3(n)$ to get $F_5(n) = F_3(n) \left(1 - \frac1{5^n} \right)$ where $F_5(n)$ has all multiples of $3$ and $5$ removed.
This can be done over all the primes. Note the adding or subtracting depends on whether the prime is $\pm 1 \mod 4$. This is due to the following reason.
If $\displaystyle (4k+1) = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} q_1^{\beta_1} q_2^{\beta_2} \cdots q_n^{\beta_n}$, where $p_i \equiv 1 \bmod 4$, $q_j \equiv 3 \bmod 4$, then $\beta_1 + \beta_2 + \cdots + \beta_n$ is even. (Why?)
If $\displaystyle (4k+3) = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} q_1^{\beta_1} q_2^{\beta_2} \cdots q_n^{\beta_n}$, where $p_i \equiv 1 \bmod 4$, $q_j \equiv 3 \bmod 4$, then $\beta_1 + \beta_2 + \cdots + \beta_n$ is odd. (Why?)
Hence, $\displaystyle \frac{1}{4k+1} = \frac1{p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} q_1^{\beta_1} q_2^{\beta_2} \cdots q_n^{\beta_n}} = \frac1{p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} (-q_1)^{\beta_1} (-q_2)^{\beta_2} \cdots (-q_n)^{\beta_n}}$
Similarly, $\displaystyle \frac{-1}{4k+3} = \frac1{p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} q_1^{\beta_1} q_2^{\beta_2} \cdots q_n^{\beta_n}} = \frac1{p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} (-q_1)^{\beta_1} (-q_2)^{\beta_2} \cdots (-q_n)^{\beta_n}}$