$\displaystyle \ln(M_{Y_n}(t))=-\frac n2 \ln\left(1-2a \frac tn-a \frac{t^2}{n}+a^2 \frac{t^3}{n^2}\right)$

Since there's a coefficient n in front of the logarithm, it is sufficient to get a $\displaystyle o\left(\frac 1n\right)$ (asymptotic equivalence)

But we know that $\displaystyle \ln(1+x)=x+o(x)$

So $\displaystyle \ln\left(1-2a \frac tn-a \frac{t^2}{n}+a^2 \frac{t^3}{n^2}\right)=-2a \frac tn-a \frac{t^2}{n}+o\left(\frac 1n\right)$ (the $\displaystyle \frac{t^3}{n^2}$ can be omitted, because the power of n is smaller than 1/n)

So $\displaystyle \ln(M_{Y_n}(t))=-\frac n2 \left(-2a \frac tn-a \frac{t^2}{n}+o\left(\frac 1n\right)\right)=at+a\frac{t^2}{2}+o(1)$

Hence, as n goes to infinity, $\displaystyle \ln(M_{Y_n}(t)) \to at+a \frac{t^2}{2}$

This gives you the limit for $\displaystyle M_{Y_n}(t)$