Well, my teacher went through a method of proving continuity of $\exp(x)$ which I don't like, so I tried to go about it a different way:
We have proved the following (which I use)
$\exp(x+y) = \exp(x)\exp(y)$
$\exp(0) = 1$
$\exp(x) \geq 1 + x \forall x \in \mathbb{R}$
firstly, I prove it's continuous at $ x = 0$
for $x \leq 0$
$\exp(-x) = \dfrac{1}{\exp(x)}$
$\exp(-x) = 1 + (-x) + \dfrac{(-x)^2}{2!} +... = 1 + $ +ve terms so $\exp(-x) \geq 1$
so $\exp(x) \leq 1$
$\Rightarrow 1 + x \leq \exp(x) \leq 1$ then by sandwich theorem $\lim_{x\to0^-}\exp(x) = 1$
for $x \geq 0 $ $-x \leq 0 \Rightarrow 1 - x \leq \exp(-x) = \dfrac{1}{\exp(x)} \leq 1$ by subbing in (-x) to the above therefore if $ 0 \leq x < 1 $ $\dfrac{1}{1-x} \geq \exp(x) \geq 1$ therefore $\lim_{x\to0^+}\exp(x) = 1$ therefore $\lim_{x\to 0} \exp(x) = 1$, I have a query at this point, can I consider $0 \leq x < 1$ and conclude $\lim_{x\to0+}\exp(x) = 1$?
Moving on, showing it is continuous for any $c \in \mathbb{R}$
assume a sequence $(x_n)$ is a seq. with $x_n \to c$ so $(x_n - c) \to 0 \Rightarrow \exp(x_n -c) \to 1$ (by the composition function theorem, and the step above)
$\exp(x_n) = \exp((x_n -c) + c) = \exp(x_n -c)\exp(c) \to \exp(c) $ how can I conclude from here?
With the axiom of choice one can prove that there exists a function $f\colon \mathbb R\to \mathbb R$ such that $f(0)=1$, $f(a+b)=f(a)f(b)$, $f(x)\ge 1+x$ and $f$ is not continuous in any point. So the assumptions you have (apart from Taylor, which you shouldn't have), are not enough to conclude the proof.
However if you have defined (as I now see in the comments) $$ \exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!} $$ you just have to notice that $$ \exp(x) = 1 + x \sum_{n=1}^\infty \frac{x^{n-1}}{n!} $$ whence for $x\to 0$ $$ |\exp(x) - 1| \le |x| \sum_{n=1}^\infty{|x|^{n-1}} \le |x| \frac{1}{1-|x|} \to 0 $$ and the continuity in $0$ is proven.
The continuity in a point $x\neq 0$ can be derived as you have described in the question.