= Solution
For the unit-rate <exponential distribution>, direct integration gives
$$
\mathbb E e^{\theta Y_1}=\frac1{1-\theta}\quad(\theta<1),\qquad \Lambda(\theta)=-\log(1-\theta).
$$
For $x>0$, maximizing $\theta x+\log(1-\theta)$ gives $\theta=1-1/x$, and therefore $I(x)=x-1-\log x$. For $x\leq0$ the supremum is infinite, as seen by taking $\theta\to-\infty$.
For $a>1$, $I$ is increasing on $[a,\infty)$, and its continuity gives $\inf_{[a,\infty)}I=\inf_{(a,\infty)}I=I(a)$. The closed-set upper bound and open-set lower bound of the <Cramér theorem> squeeze the desired limit to $-I(a)$. For $a=1$, both infima are $0$, because $I(1)=0$ and $I(x)\to0$ as $x\downarrow1$. For $0\leq a<1$, the <strong law of large numbers> gives $S_n/n\to1$ almost surely, so the tail <probability> tends to $1$; at $a=0$ it is identically $1$.
Consequently the <exponential sample-mean upper-tail rate> is
$$
\boxed{\lim_{n\to\infty}\frac1n\log\mathbb P(S_n\geq na)=\begin{cases}0,&0\leq a\leq1,\\-(a-1-\log a),&a>1.\end{cases}}
$$
The threshold case uses the rate-function lower bound on the open half-line, rather than applying the law of large numbers at equality.
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