There is no homology theory which satisfies the following conditions:

$H_0(X)$ is the free abelian group generated by the connected
components of $X$.

The homomorphism $f_*:H_0(X)\to H_0(Y)$ induced by a continuous map
$f:X\to Y$, maps a generator $[x]\in H_0(X)$ to the generator $[f(x)]\in
H_0(Y)$.

The theory satisfies the Homotopy, Exactness, Excision and Dimension
axioms.

Proof. Let
$$X=\{(0,1)\} \cup ([0,1]\times \{0\}) \cup \bigcup\limits_{n\ge
1} (\{\frac{1}{n}\}\times [0,1]) \subseteq \mathbb{R}^2$$
with the usual subspace topology. Let $A=X\smallsetminus \{(0,1)\}$ and $U=\{(x,y)\in X \ | \ y<x\} \subseteq X$. Then $U$ and $A$ are open and
$\overline{U}\subseteq int(A)=A$. By Excision $H_0(X\smallsetminus U,
A\smallsetminus U)$ is isomorphic to $H_0(X,A)$. Since $A$ and $X$ are
connected, $H_0(A)\to H_0(X)$ is the identity by 2. Since $A$ is
contractible, by Homotopy it has the homology of a point, so by Dimension
$H_{-1}(A)=0$. Then by Exactness $H_0(X,A)=0$. On the other hand
$\{(0,1)\}$ is a connected component of $X\smallsetminus U$ and
$(0,1)\notin A\smallsetminus U$. Therefore $H_0(A\smallsetminus U)\to
H_0(X\smallsetminus U)$ is not surjective (by 1 and 2). By Exactness,
$H_0(X\smallsetminus U)\to H_0(X\smallsetminus U, A\smallsetminus U)$ is
not the trivial homomorphism. Then $H_0(X\smallsetminus U, A\smallsetminus
U)\neq 0$, a contradiction.