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Let $X$ be a topological space and $A_i$ abelian groups indexed over some (possibly infinite) set $I$.

Can one express the singular homology $H_\ast(X,\bigoplus_{i\in I}A_i)$ in terms of $H_*(X,A_i)$? Is it just $\bigoplus_{i\in I}H_\ast(X,A_i)$?

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    Coefficient should be a ring, not an abelian group. In that case singular homology becomes a module over the ring of coefficient. Obviously you cannot sum modules over different rings. So this is doesn't make sense if you look at it strictly. – 2017-01-27
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    You can certainly use abelian groups as coefficients -- then homology has just an abelian group structure. – 2017-01-27
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    @Pedro How do you define homology over an abelian group? One of the steps (chain complex) is to define a free module over a ring. How do you do that if the underlying coefficients is only an abelian group, not a ring? – 2017-01-27
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    Elements of the chain complex are given by formal sums of singular simplices, where the coefficients belong to your given abelian group. You then get an abelian group structure on the chain complex. You define the boundary map as usual, and you get homology as an abelian group. You can find this in Hatcher. – 2017-01-27
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    Every abelian group is a module over the integers. – 2017-01-27

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To answer the question in the comments, if $C_\bullet (X)$ is the usual singular complex formed by free abelian groups spanned by singular simplices, one defines $H_\bullet (X;A)$ to be the homology of the complex of abelian groups $C_\bullet (X)\otimes_\mathbb{Z} A$.

Now note that tensor products commute with direct sums: $$C_\bullet (X)\otimes_\mathbb{Z} \left(\bigoplus_i A_i\right) \cong \bigoplus_i C_\bullet (X)\otimes_\mathbb{Z} A_i,$$ and that taking homology commutes with direct sums as well: $$H_n (\bigoplus_i C_\bullet^{(i)}) \cong \bigoplus_i H_n (C_\bullet^{(i)}).$$ (This is true for complexes of abelian groups, or $R$-modules over any $R$; in general, this is related to Grothendieck's axiom AB4.)