Is this Epsilon-Delta approach to prove that $e^x$ is continuous correct?

It looks okay, although there are a few things one can do to it. You also have to be careful with taking logs if $e^{x_0}-\epsilon<0$...

I would suggest the following: you have $$ \lvert e^x-e^{x_0} \rvert = e^{x_0} \lvert e^{x-x_0}-1 \rvert < \varepsilon $$

Now, you have the elementary inequality $$ e^y \geqslant 1+y, $$ which is easy to prove with even the definition as $\lim_{n \to \infty} (1+y/n)^n$. Substituting $y \to -y$, you have $$ e^{-y} \geqslant 1-y, $$ and then for $y<1$, both sides are positive, so dividing gives $$ \frac{1}{1-y} \geqslant e^y, \quad (y<1) $$ Hence $$ \lvert e^y-1 \rvert \leqslant \max{\left\{|y|,\left\lvert \frac{y}{1-y} \right\rvert\right\}}, \quad (y<1) $$ and then all you have to do is take $\delta$ small enough that $y=x-x_0$ satisfies $$ \max{\left\{|y|,\left\lvert \frac{y}{1-y} \right\rvert\right\}}<e^{-x_0}\varepsilon $$ and $|y|<1$.

(Oh, and this also shows that the exponential is not uniformly continuous, since you can't get rid of the $x_0$-dependence in a uniform way)

One nice thing about $e^x$ is that it has a scale invariance property which reduces continuity everywhere to continuity at any particular point. Suppose $e^x$ is continuous at $0$. Then

$$|e^x-e^{x_0}|=|e^{x-x_0}e^{x_0}-e^{x_0}|=e^{x_0}|e^{x-x_0}-1|<\epsilon$$

if we choose $\delta$ such that $|e^y-1|<\epsilon/e^{x_0}$ for all $|y|<\delta$ (and substituting $y=x-x_0$ to get the above).

To show continuity at $0$, there are different approaches depending on your chosen definition. For the infinite series definition, we can use the squeeze theorem to show $1\le e^x\le\frac1{1-x}$ for $x\in (0,1)$:

$$\frac{x^0}{0!}\le\sum_{n=0}^\infty\frac{x^n}{n!}\le\sum_{n=0}^\infty x^n.$$

And for $x<0$ we can substitute $x\mapsto-x$ to get $1\le e^{-x}\le\frac1{1+x}\implies 1+x\le e^x\le1$ (by taking the reciprocal). (In case it wasn't clear, this gives an explicit $\delta$ to use; from the inequalities it follows that $\delta=\min(1,\epsilon,\frac{\epsilon}{1+\epsilon})=\frac{\epsilon}{1+\epsilon}$ is sufficient.)