In order to prove this, I had to back the proof up into a number of Lemmas. I am wondering if my method was reasonable, or if there was a more general or expedient path I might have taken? (I have left out all of the actual proofs with the exception of Lemma 6.)
Lemma 1:
For any $a\in\mathbb{R}$, $$f(a)=1+a \implies f(-a)=1-a.$$
Lemma 2:
For any $n\in\mathbb{N}$, $\quad f(n)=1+n$.
Lemma 3:
For any $n\in\mathbb{N}$, $\quad f\left(\frac{1}{n}\right)=1+\frac{1}{n}$.
Lemma 4:
For any rational $\frac{m}{n}$, such that $0<m<n$, $\quad f\left(\frac{m}{n}\right)=1+\frac{m}{n}$.
Lemma 5:
For any $a\in\mathbb{Q}$, $\quad f(a)=1+a$.
Lemma 6:
For any irrational $x, \quad f(x)=1+x.$
Proof: Assume, for some irrational $x, \quad f(x)\neq 1+x$.
Say $f(x)>1+x$, then $$f(x)-(1+x)=t>0$$
and so, for any rational $a<x$ $$ f(x)-(1+a)>t.$$
However, by the definition of continuity, with $\epsilon=t$, for any rational satisfying $$x-\delta<a<x \quad\implies f(x)-(1+a)<t,$$
which is a contradiction. A similar argument shows that assuming $f(x)<1+x$ also leads to a contradiction, and it follows that $$f(x)=1+x.\quad \square$$
If you make a translation $g(x) = f(x-1)$ the problem reduces to finding all continuous functions $g$, such that $g(1) = 1$, $g(x+y) = g(x) + g(y)$ for all real $x, y$. The second of these makes $g$ what is called an additive function; is a very standard functional equation which leads to $g(x)= g(1)x$ when any of the following additional assumptions is satisfied:
$g$ is continuous at some point $x_0$ (you have that it is continuous everywhere, but it would have been enough to know it is continuous at any one arbitrary point)
$g$ is bounded on some interval
$g$ is monotonic on some interval
The proof, in general, proceeds exactly along the lines of those lemmas. But if you are allowed to use that a continuous additive function is always $g(x) = g(1)x$ as a theorem, you can "skip" all the lemmas.