The point is that by your construction $f:\mathcal{E}_\Gamma\to Y_\Gamma$ is an elliptic surface with a section $\sigma$. So as Keerthi said in the comments, $f$ is projective. In fact the relative divisor $3\sigma$ gives the Weirstrass embedding $\mathcal{E}_\Gamma\to \mathbb{P}(f_*\mathcal{O}(3\sigma))$). (Ben's comment shows that projectivity can fail without a section.) From here, quasiprojectivity is straight forward: If $V$ is an extension of $f_*\mathcal{O}(3\sigma)$ to a vector bundle on the smooth projective closure of $ Y_\Gamma$, then the closure of $\mathcal{E}_\Gamma$ in $\mathbb{P}(V)$ is a projective variety. Therefore $\mathcal{E}_\Gamma$ is quasiprojective.
Given the endless stream of comments, perhaps I should add a few words of clarification:
- Since $Y=Y_\Gamma$ is noncompact, any vector bundle such as $f_*\mathcal{O}(3\sigma)$ is in fact (analytically) trivial. This is definitely overkill, but you can use Grauert, "Analytichse Faserungen..." Math. Ann 1958.
- Thus $\mathbb{P}(\mathcal{O}(3\sigma))\cong Y\times \mathbb{P}^2$. This embeds into $\bar{Y}\times \mathbb{P}^2$, where $\bar Y$ is a (the) smooth projective compactification of $Y$.
- We can take the closure of $\mathcal{E}_\Gamma$ to get a projective variety, and the quasiprojectivity of this family follows easily.
- Note that the fibres of the closure $\overline{\mathcal{E}_\Gamma}$ may be singular.
- If the Hopf surface had a section, it would lift to a rational curve in $\mathbb{C}^2-\lbrace 0\rbrace$. Well, I'll let you think about why that might be a problem.