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.

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Given the endless stream of comments, perhaps I should add a few words of clarification:

 1. 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.
 2. 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 $Y$.
 3. We can take the closure of $\mathcal{E}_\Gamma$ to get a projective variety, and the quasiprojectivity of this family follows easily.
 4. Note that the fibres of the closure $\overline{\mathcal{E}_\Gamma}$ may be singular.
 5. 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.