Your question is no longer impossible. As people mentioned in the comments, there are many different spaces with $\pi_1(X)\cong A_5$, but your question is now framed in a way that suggests looking at a particular such space.
Using the presentation $A_5\cong\langle a,b\mid a^2,b^3,(ab)^5\rangle$ given in the problem statement, one can construct a 2-dimensional CW complex $X$ with $\pi_1(X)\cong A_5$ as follows: give $X$ a single $0$-cell $e^0=*$, 2 different $1$-cells $e^1_a,e^1_b$ with attaching maps $\partial e^1_a,\partial e^1_b\to X^0=\{*\}$ necessarily collapsing their boundaries to a point (i.e. the 1-skeleton $X^1$ is just a wedge of two circles), and 3 different $2$-cells $e^2_{a^2},e^2_{b^3},e^2_{(ab)^5}$ with attaching maps described by their subscript.
For example, the attaching map $S^1\cong\partial e^2_{(ab)^5}\to X^1\cong S^1\vee S^1$ corresponds to the element $$(ab)^5\in\pi_1(X^1)\cong\langle a,b\rangle$$ (i.e. its a loop that goes around the $e^1_a$ cell and then goes around the $e^1_b$ cell and then around the $e^1_a$ cell again and then around the $e^1_b$ cell again and so on 5 times in total).
Using Van Kampen, one shows that this space $X$ has $A_5\cong\langle a,b\mid a^2,b^3,(ab)^5\rangle$ as its fundamental group.
We now turn to the homology of $X$. As Mindlack mentioned, since $A_5$ is simple (and nonabelian), $H_1(X)\cong\pi_1^{\mathrm{ab}}(X)\cong A_5/[A_5,A_5]=0$. Since $X$ is a 2-dimensional CW-complex we also know $H_k(X)=0$ for all $k\ge3$, so we're just left with $H_2(X)$. For this, we use the cellular chain complex
$$0\to H_2(X^2,X^1)\to H_1(X^1, X^0)\to H_0(X^0)\to 0,$$
which, in this case, looks like
$$0\to\mathbb Z^{\oplus 3}\to\mathbb Z^{\oplus 2}\to\mathbb Z\to 0,$$
so we're interested in calculating the map $\mathbb Z^{\oplus 3}\to\mathbb Z^{\oplus 2}$ above. This map sends $[e^2_{a^2}]\mapsto c[e^1_a]+d[e^1_b]$ where $c$ is the degree of the map
$$S^1\cong\partial e^2_{a^2}\to X^1\to X^1/(X^1\setminus e_a^1)\cong S^1$$
and $d$ is the degree of the map
$$S^1\cong\partial e^2_{a^2}\to X^1\to X^1/(X^1\setminus e_b^1)\cong S^1.$$
Since $e^2_{a^2}$ is attached via the word $a^2$ (i.e. it loops around $e_a^1$ twice), we see that $c=2$ and $d=0$.
One can similarly determine the images of $e^2_{b^3}$ and $e^2_{(ab)^5}$. In the end, you get that the map $\mathbb Z^{\oplus 3}\to\mathbb Z^{\oplus 2}$ is given, using the ordered bases $\{e^2_{a^2},e^2_{b^3},e^2_{(ab)^5}\}$ and $\{e^1_a,e^1_b\}$, by the matrix
$$\begin{pmatrix}
2 & 0 & 5\\
0 & 3 & 5
\end{pmatrix}$$
From the cellular chain complex, the second homology group $H_2(X)$ is given by the kernel of this matrix, so we see that
$$H_2(X)\cong\mathbb Z$$
with explicit generator $15[e^2_{a^2}]+10[e^2_{b^2}]-6[e^2_{(ab)^5}]$$.
