Continuity proof.

There's a sense of deja vu here. I'm not exactly what you consider axiomatic.

Here's a different approach for $\exp$.

The function $x \mapsto \exp(x)$ is, by your definition, the inverse of $x \mapsto \log(x)$. The function $\log$ is continuous, strictly increasing, and the range is $\mathbb{R}$. Hence it follows that $\exp$ is strictly increasing and defined on all of $\mathbb{R}$. Let $\epsilon>0$, and let $\delta = \min\{-\log(1-\epsilon), \log(1+\epsilon)\}$, and suppose $|x| < \delta$. Then since $$\log(1-\epsilon) < x < \log(1+\epsilon)$$ we have $$ 1-\epsilon = \exp(\log(1-\epsilon)) < \exp(x) < \exp(\log(1+\epsilon)) = 1+\epsilon$$ Hence $|\exp(x)-1| < \epsilon$, and so $\exp$ is continuous at $0$.

Original 'non-answer':

For $\sin$, you know that $\sin^2(x)+\cos^2(x) = 1$, hence $\cos(x)$ is bounded by $1$. Since $\cos$ is the derivative of $\sin$, the mean value theorem shows that $|\sin(x)-\sin(y)| \leq |x-y|$ for any $x,y$.

Replacement answer:

In context of the definition of $\sin$ from the question: The function $x \mapsto A(x)$ is strictly increasing and continuous on $[-1,1]$. Using a similar argument to that used for $\exp$ above, the function $\cos$ can be seen to be continuous. Hence, since $\sin(x) = \sqrt{1-\cos^2(x)}$, it follows that $\sin$ is continuous.