This problem is incorrectly stated. The Final Value Theorem is stated as follows (or check out a signals and systems text):
If $f$ is bounded on $(0,\infty)$ and $\lim\limits_{t\rightarrow\infty}f(t) < M$ for some $M$, then
\begin{align*}
\lim\limits_{t\rightarrow\infty}f(t) = \lim\limits_{s \searrow 0} sF(s)
\end{align*}
where $F(s)$ is the unilateral Laplace Transform of $f$.
With this, the final value (DC Gain) is $0$. The OP has this in their solution, since $\lim\limits_{t\rightarrow \infty} e^{-at} \rightarrow 0$, assuming $a>0$. Otherwise, the function is unbounded, and the final value theorem does not apply since it has a pole outside of the open left half plane.
Here's what I think the problem meant to say: Let $a>0$ and
$Y(s) = G(s)U(s) = \frac{1}{s}\frac{A}{s+a}$. Using the FVT, find the DC gain, and the time domain response.
Then, applying FVT gives,
\begin{align*}
\lim\limits_{s \searrow 0} sY(s) = \lim\limits_{s \searrow 0} s\frac{A}{s+a}\frac{1}{s} = \lim\limits_{s \searrow 0} \frac{A}{s+a} = \frac{A}{a}
\end{align*}
As the solution above suggests, partial fractions can be used to get the total response, or, if you have been introduced to some complex analysis, you can also use Cauchy's Residue Theorem.
Applying partial fractions should give $\frac{A/a}{s} + \frac{-A/a}{s+a}$, which when factored gives the final answer. Note that if you take $t\rightarrow \infty$, you find the final value (DC gain) as expected.
They are both correct when $F$ is a rational function or the product of a rational with an exponential. The original statement of the theorem postulates that $f$ is bounded and $\lim_{t\to\infty} f(t)$ exists. The question is how we can know that given $F$.
For $f$ to be bounded and for the limit $\lim_{t\to\infty} f(t)$ to exist, the (rational) function $F$ must have its poles in the open left half plane.
Moreover, $F$ is guaranteed to exist because $f$ is bounded. There is a little caveat here: We have
$$
\int_0^\infty f(t)e^{-st}\mathrm{d}t
\leq
\int_0^\infty |f(t)|e^{-st}\mathrm{d}t
\leq
\int_0^\infty Me^{-st}\mathrm{d}t ,
$$
which converges for all $s \in \mathbb{C}$ with $\Re(s) > 0$. However, we must assume that $f(t)e^{-st}$ is integrable. It suffices to assume that $f$ has a finite number of discontinuities, or assume that the integral is meant in the Lebesgue sense.
Best Answer
Functions with Laplace transforms that have poles in the right half plane don't have a final value. Take $e^t-1$, for example. Its Laplace transform is
$$\frac{1}{s-1}-\frac{1}{s}= \frac{1}{s(s-1)}$$
and the Final Value Theorem would have you believe that
$$e^{\infty} - 1 = -1$$
which is clearly absurd.