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Since

$ G =( prod_1^infty mathbbZ )times mathbbZ times mathbbZ cong (prod_1^infty mathbbZ) times mathbbZ $,

by taking quotients we get

$mathbbZ times Z cong G/ prod_1^infty mathbbZ cong mathbbZ$.

Therefore

$mathbbZ cong mathbbZ times Z$.

But $mathbbZ$ is indecomposable! What's wrong with above proof?

Quotienting by isomorphic copy need not preserve isomorphisms. For a simpler related example: $C_n=mathbbZ/nmathbbZ$ and $mmathbbZcong nmathbbZ$ for all $m,n>0$, but obviously the cyclic groups $C_m$ and $C_n$ are not isomorphic if $mneq n$.

Your argument is based on the assumption that $G/N_1cong G/N_2$ follows from $N_1cong N_2$. Here is a simple example to show that this assumption is incorrect.

The non-isomorphic groups $mathbb Z_4$ and $mathbb Z_2timesmathbb Z_2$ are both homomorphic images of the group $G=mathbb Z_2timesmathbb Z_4$. Thus there are subgroups $N_1,N_2$ of $G$ such that $G/N_1congmathbb Z_4$ and $G/N_2congmathbb Z_2timesmathbb Z_2$. The subgroups $N_1$ and $N_2$ are isomorphic, as both have order $2$; so $N_1cong N_2$ but $G/N_1notcong G/N_2$.