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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.