Write each of the signals as
$$e^{-k \tau} \theta(\tau)$$
where $k$ is either of $a$ or $b$, and $\theta(\tau)$ is the Heaviside step function, zero when $\tau < 0$ and $1$ when $\tau > 0$. The convolution integral may then be written as
$$\int_{-\infty}^{\infty} d\tau \, e^{-a \tau} \theta(\tau) \, e^{-b (t-\tau)} \theta(t-\tau)$$
Now, the product of the two Heavisides in the integral is zero outside the interval $[0,t]$. Therefore, we may write the convolution integral as
$$\int_0^t d\tau \, e^{-a \tau} \, e^{-b (t-\tau)} = e^{-b t} \int_0^t d\tau \, e^{-(a-b) \tau} $$
which is
$$\frac{1}{a-b} e^{-b t} \left (1-e^{-(a-b) t} \right ) = \frac{e^{-b t}-e^{-a t}}{a-b}$$
Yes, the convolution of an integrable function $f$ with compact support, and a Schwartz class function $g$ belongs to the Schwartz space again.
Since all derivatives of Schwartz class functions belong to the Schwartz space, in particular are bounded, the convolution
$$(f\ast g)(x) = \int f(y)g(x-y)\,dy$$
is smooth, since the dominated convergence theorem allows differentiating under the integral arbitrarily often. (Since the difference quotients of $\partial^\alpha g$ converge uniformly on $\mathbb{R}$, and the support of $f$ is compact, one can get that result also without the dominated convergence theorem.)
So only the decay remains to be checked. Choose $K > 0$ such that $\operatorname{supp} f \subset [-K,K]$. Since $g\in \mathcal{S}(\mathbb{R})$, for every $\alpha,m\in\mathbb{N}$ there is a constant $C_{\alpha,m}$ such that
$$\lvert \partial^\alpha g(x)\rvert \leqslant \frac{C_{\alpha,m}}{(1+\lvert x\rvert)^m}$$
for all $x\in\mathbb{R}$.
Then for $\lvert x\rvert \geqslant 2K$ we have
$$\begin{align}
\lvert \partial^\alpha(f\ast g)(x)\rvert &= \left\lvert \int_{-K}^K f(y) \partial^\alpha g(x-y)\,dy \right\rvert\\
&\leqslant \int_{-K}^K \lvert f(y)\rvert\, \lvert \partial^\alpha g(x-y)\rvert\,dy\\
&\leqslant \int_{-K}^K \lvert f(y)\rvert \frac{C_{\alpha,m}}{(1+\lvert x-y\rvert)^m}\,dy\\
&\leqslant \int_{-K}^K \lvert f(y)\rvert \frac{C_{\alpha,m}}{\left(1 + \frac{\lvert x\rvert}{2}\right)^m}\,dy\\
&= \frac{2^mC_{\alpha,m}}{(2+\lvert x\rvert)^m}\int_{-K}^K\lvert f(y)\rvert\,dy\\
&\leqslant \frac{C'_{\alpha,m}}{(1+\lvert x\rvert)^m},
\end{align}$$
where $C'_{\alpha,m} = 2^m\lVert f\rVert_{L^1}C_{\alpha,m}$.
Since $(1+\lvert x\rvert)^m \partial^\alpha(f\ast g)(x)$ is continuous, it is bounded on the compact set $[-K,K]$, hence we have
$$(1+\lvert x\rvert)^m\lvert\partial^\alpha(f\ast g)(x)\rvert \leqslant \tilde{C}_{\alpha,m}$$
for all $x\in\mathbb{R}$ and some constant $\tilde{C}_{\alpha,m}$.
So $f\ast g$ is a smooth function such that $x^m\partial^\alpha(f\ast g)(x)$ is bounded for all $\alpha,m\in\mathbb{N}$, and that means precisely $f\ast g\in \mathcal{S}(\mathbb{R})$.
The generalisation to $\mathbb{R}^n$ is immediate.
Best Answer
Definition: Let $f:\mathbb{R}^N\to\mathbb{R}$ be a function. Consider the family $\omega_i$ of all open sets on $\mathbb{R}^N$ such that for each $i$, $f=0$ a.e. in $\omega_i$, then $f=0$ a.e. on $\omega=\cup\omega_i$ and we define the support of $f$ by $$\operatorname{supp}(f)=\mathbb{R}^N\setminus\omega$$
Suppose that $f,g\in C^\infty(\mathbb{R})\cap L^1(\mathbb{R})$. We can prove that $\operatorname{supp}(f\ast g)\subset\overline{\operatorname{supp}(f)+\operatorname{supp}(g)}$.
Take any $x\in\mathbb{R}$ and note that $$(f\ast g)(x)=\int_\mathbb{R}f(x-y)g(y)=\int_{(x-\operatorname{supp}(f))\cap \operatorname{supp}(g)}f(x-y)g(y)$$
For $x\notin \operatorname{supp}(f)+\operatorname{supp}(g)$, we have that $(x-\operatorname{supp}(f))\cap \operatorname{supp}(g)=\emptyset$ and then $(f\ast g)(x)=0$. It follows that if $x\in (\operatorname{supp}(f) + \operatorname{supp}(g))^c$ then $$(f\ast g)(x)=0$$
Therefore $$\operatorname{supp}(f\ast g)\subset\overline{\operatorname{supp}(f)+\operatorname{supp}(g)}$$
Now let's prove that the equality is not necessary. Consider the functions in $\mathbb{R}$ defined by $f(x)=x\chi_{[-1,1]}(x)$ and $g(x)=\chi_{[-2,2]}(x)$, where $\chi_A$ is the characteristic function of the set $A$.
Note that in this case $\overline{\operatorname{supp}(f)+\operatorname{supp}( g)}=[-3,3]$, however, the interval $(-1,1)$ does not belong to the support of $f\star g$ (which is equal to $[-3,-1]\cup [1,3]$).
Remark 1: The proof I gave here can be found in Brezis's book of Functional Analysis.
Remark 2: I don't know a expliclty characterization of the support of the convolution, but by the given formula, you can see that if the two functions has compact support, then does the convolution.
Update: I have corrected some errors in the text.