Yes, $\sum_{n=1}^\infty f_n= f$ is a sum of functions and is defined by saying that, for a specific value of x, f(x) is the value of the numeric sum $\sum_{n=1}^\infty f_n(x)$.
"And we specify whether we mean uniformly or pointwise". Not quite! Convergence of a sequence of series of functions is always "pointwise". It may then "converge uniformly" or not. The definition of convergence of "$\lim_{n\to\infty} f_n= f$" is that the numerical sequence "$\lim_{n\to\infty} f_n(x)$" converges to "$f(x)$" for every x in the domains of all the $f_n$. That is "pointwise" convergence- it converges at every "point"- every x-value.
We can then determine whether or not that convergence is also uniform.
Determining "pointwise" convergence is exactly the same as for numerical sequence. A sequence of functions, $f_n$, converges pointwise to f if and only if the numerical sequence, $f_n(x)$, converges to f(x) for every x in the domain. That is, if for a specific x, given $\epsilon> 0$ there exist N such that if n> N then $|f_n(x)- f(x)|< \epsilon$. This is done at every value of x- given an $\epsilon$, what N will work may be different for different x values.
The sequence of functions converges uniformly (on some interval) if, in addition to converging pointwise, given $\epsilon$ there exist an N that will work for all x in that interval.
Notice I was talking about sequences here. A series, $\sum_{n= 1}^\infty f$ converges to f if and only if the sequence of finite sums, $\sum{n=1}^N f_n$ converges to f so the same distinction between "pointwise" and "uniform" convergence applies. A series, if it converges, always converges "pointwise". It them may or may not converge "uniformly".
Best Answer
You can directly show that the power series for $f$ satisfies $f(x+y) = f(x)f(y)$. Note that the coefficient of $x^iy^j$ on both sides equals $\frac{1}{i!j!} = \frac{1}{(i+j)!}\binom{i+j}{i}$. This property is saying that $f$ is a homomorphism from the additive reals onto the multiplicative group of nonzero reals.
It is straightforward to check by induction that this guarantees that $f(n) = f(1)^n$ and $f(1/n) = f(1)^{1/n}$. More generally, we have that $f(q) = f(1)^q$ for all rational numbers $q$. Since $f(x)$ is continuous, it therefore must be equal to $f(1)^x$.