It is true that statements like this are completely classical and yet hard to find stated just like that in the literature! If noone has a reference, I would suggest to go as follows :

-- the cohomology groups are given in the book by Cartan and Eilenberg (oldest reference I can think of!)

-- the cohomology ring is obtained by looking at the exact sequence $G \to S^1 \to S^1$ where $G$ is your cyclic group of order $n$ and the map $S^1 \to S^1$ is multiplication by $n$; there is a fibration $G \to S^1 \to S^1 \to BG \to BS^1 \to BS^1$, and the "Gysin exact sequence" of the $S^1 \to BG \to BS^1$ part gives you the multiplicative structure. You have to use that $BS^1 = \mathbb{P}^1(\mathbb{C})$. In a nutshell: it is standard algebraic topology.

--this next step is probably more relevant to your question. It is always the case that $H^1(G, \mathbb{F}_2) = Hom(G, \mathbb{F}_2)$ (reference: any book on cohomology, such as Cartan-Eilenberg). This identifies the elements in degree $1$ as Stiefel-Whitney classes by definition (of course $\mathbb{F_2} = O_1(\mathbb{R})$...)

Then, the exponential exact sequence $0\to \mathbb{Z} \to \mathbb{C} \to \mathbb{C}^\times \to 1$ gives you $H^2(G, \mathbb{Z}) = Hom(G, S^1) ~~~(= Hom(G, GL_1(\mathbb{C})))$ when $G$ is finite. From there you can identify the elements in degree $2$.

OK this is way too long. I agree that a good reference would be better.