Homology of a finite graph follows from Mayer-Vietoris sequence?

$\newcommand{\Z}{\mathbb{Z}}$Here's a proof without reduced homology (which, by the way, makes the problem much simpler). To sum up what you've written (as far as I can tell it's correct), you're left with:

$$0 \to \underbrace{H_1(U) \oplus H_1(V)}_{\Z^{e-v}} \to H_1(X) \to \underbrace{H_0(U \cap V)}_{\Z^2} \xrightarrow{g} \underbrace{H_0(U) \oplus H_0(V)}_{\Z^2} \xrightarrow{f} \underbrace{H_0(X)}_{\Z} \to 0$$

$f$ sends $(a,b)$ to $a-b$ (because this is the MV sequence). The map $g$ sends both $(1,0)$ and $(0,1)$ to $(1,1)$ (easy examination). Finally you're left with an exact sequence:

$$0 \to \Z^{e-v} \to H_1(X) \to \Z \to 0$$ where the last $\Z$ is generated by $(1,-1)$ in $H_0(U \cap V) \simeq \Z^2$. By general facts about exact sequences (namely, $\Z$ is projective because it is free), you get $H_1(X) \simeq \Z^{e-v} \oplus \Z$.