Can I get $e^x$ from the Taylor series of $e$? If I start with the series $\sum_{n=0}^{\infty} \frac{1}{n!}$ and call it $e$, can I show that $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ is $e^x$ using only this series?

Sure.  If you define
$$F(x) = 1 + x + x^2/2! + \cdots$$
and define $$e = F(1) = 1 + 1 + 1/2! + \cdots$$ your goal is to show that $F(x) = F(1)^x$ for all real $x$.
You can see quite easily that $F$ is its own derivative, and $F(0) = 1$.
You can therefore argue that $F(n) = F(1)^n$ for all natural numbers $n$ by showing that the derivative of the quotient function $\frac{F(nx)}{F(x)^n}$ is zero, concluding that $\frac{F(nx)}{F(x)^n}$ must be constant, and then showing that this constant must be $1$ by substituting $x=0$.
It is not difficult to argue more generally that $F(a) = F(1)^a$ for any rational number $a$.  It follows by continuity that $F(x) = F(1)^x$ for all real $x$, which establishes what you want, namely that
$$1 + x + x^2/2! + \cdots = F(x) = F(1)^x = e^x.$$
