Have $k$ be a positive integer such that $(x_1,y_1)$ and $(x_2,y_2)$ are two distinct solutions in non-negative integers, so
$$ax_1+by_1=k=ax_2+by_2 \;\;\to\;\; a(x_1-x_2)=b(y_2-y_1) \tag{1}\label{eq1A}$$
Since $\gcd(a,b)=1$, then
$$a \mid y_2-y_1 \;\;\to\;\; y_2 - y_1 = ma, \;\; m\in\mathbb{Z}, \;\; m \neq 0 \tag{2}\label{eq2A}$$
If $m \gt 0$, then since $y_1 \ge 0$, we have $y_2 \ge ma \;\;\to\;\; y_2 \ge a$. Using this in \eqref{eq1A}, since $x_2 \ge 0$, gives that
$$k = ax_2 + by_2 \ge ab \tag{3}\label{eq3A}$$
Similarly, if $m \lt 0$, we get from \eqref{eq2A} that $y_1 - y_2 = -ma$ so $y_1 \ge a$, i.e., the indices are just switched around. Thus, in either case, we get from \eqref{eq3A} that $k \ge ab$. However, as you already stated, $k \le ab$ (since $(0,a)$ and $(b,0)$ are $2$ solutions for $k = ab$), this means
$$k = ab \tag{4}\label{eq4A}$$
is the smallest such $k$.
For your more general case where $n \ge 3$, I don't offhand know of any simple solution. Nonetheless, to get a generally quite reasonable bound, from among $1 \le i \lt j \le n$, choose the index values giving the minimum value of $\operatorname{lcm}(a_{i},a_{j}) = \frac{a_{i}a_{j}}{\gcd(a_i,a_j)}$. Then have
$$d = \gcd(a_i,a_j), \;\; a_i = dc_i, \;\; a_j=dc_j, \;\; \gcd(c_i,c_j)=1 \tag{5}\label{eq5A}$$
With $x_m = 0$ for all $m \neq i,j$, then $k = dc_ic_j$ (i.e., $\operatorname{lcm}(a_{i},a_{j})$) has $2$ solutions with $(x_i,x_j)$ being $(0,c_i)$ and $(c_j,0)$.
