I'm trying to prove that if $$ f ( x + n ) = f ( x ) f ( n ) $$ for all $ x \in \mathbb R $ and $ n \in \mathbb N $, then it also holds for $ x , n \in \mathbb R $. One "argument" I came up with was regarding the symmetry. There's no reason why one should be constrained to the integers, while the other can be any real number, but that's not a proper argument.
Another thing I thought of is this: If we set $ n = 1 $ then we get $$ f ( x + 1 ) = f ( 1 ) f ( x ) \tag 1 \label 1 $$ which is true for all $ x \in \mathbb R $. Now, if we instead set $ x = 1 $ then we get $ f ( n + 1 ) = f ( n ) f ( 1 ) $ which must also be true for $ n \in \mathbb R $ because \eqref{1} is. What keeps me from being satisfied is that $ n \in \mathbb R $ under the assumption that $ x = 1 $.
Is my reasoning valid or not?
Edit: Forgot an important bit of information: $ f $ is a continuous function.
Your proof is not valid; you are noticing that if $x$ is an integer then $n$ could vary anywhere in the reals, but this is just stating that the relation $f(x+n)=f(x)f(n)$ has a symmetry where we swap $x$ and $n$ and doesn't tell us anything about whether $f(x+y)=f(x)f(y)$ when both are real.
More directly, the statement you're trying to prove is false, so obviously your proof of it is not valid. Let $f$ be the function $$f(x)=a^{x+g(x)}$$ where $g(x)$ is a function such that $g(x+1)=g(x)$ and $g(0)=0$, and clearly, for any integer $n$, it holds that $g(x+n)=g(x)$ and $g(n)=0$. Then, for integer $n$ and real $x$, it olds that $$f(x+n)=a^{x+n+g(x+n)}$$ but we can write $x+n+g(x+n)=(x+g(x))+(n+g(n))$ so we have
$$f(x+n)=a^{x+g(x)}a^{n+g(n)}=f(x)f(n).$$ However, it's easy to choose reals for which $f(x+y)=f(x)f(y)$ does not hold; for instance, choosing $x=y=\frac{1}2$ and letting $k=g(\frac{1}2)+\frac{1}2$, we get $$f(1)=f\left(\frac{1}2\right)f\left(\frac{1}2\right)$$ $$a = a^k\cdot a^k = a^{2k}$$ which does not hold if $a\neq 1$ and $g(\frac{1}2)\neq 0$.