You can look at the length of words.
Any element of $A^+$ is a concatenation
$$ a_1 a_2 \ldots a_n $$
where $a_i \in A$ and $n \geq 1$. The length of such a word is
$$ L(a_1 a_2 \ldots a_n) = L(a_1) + L(a_2) + \ldots + L(a_n) $$
But the length is always nonnegative; for example,
$$ 0 \leq L(a_1) \leq L(a_1) + L(a_2) + \ldots + L(a_n)$$
If you're given that
$$ \epsilon = a_1 a_2 \ldots a_n $$
then putting all of the above together gives
$$ 0 \leq L(a_1) \leq L(\epsilon) = 0 $$
and thus $L(a_1) = 0$, so $a_1 = \epsilon$ and $\epsilon \in A$.
If you were to assume that $\epsilon \notin A$, you could modify the above argument by the fact that $L(a_i)$ must be positive, and so
$$ 0 < L(a_1) \leq L(a_1) + L(a_2) + \ldots + L(a_n) $$
and then the final deduction would be
$$ 0 < L(\epsilon) $$
which is a contradiction.