Can every regular expression be written as sum of products?
I think you have to be more careful in the inductive steps, but it seems you basically have the argumentation ready. According to me proper induction would be


  If $R_1$ and $R_2$ can be written as sum of products, then so can
  $R_1\cdot R_2$, $R_1+R_2$ and $R_1^*$.


The proof could be completed as:

If $R_1$ and $R_2$ can be represented as sum of products as $R_1 = R_{11} + R_{12} + ... + R_{1n}$ and $R_2 = R_{21} + R_{22} + ... + R_{2m}$, then:


$R = R_1\cdot R_2$ will be equivalent to ($R_{11}\cdot R_{21} + R_{11}\cdot R_{22} + ... + R_{11}\cdot R_{2m}) + ... (R_{1n}\cdot R_{21} + R_{1n}\cdot R_{22} + ... + R_{1n}\cdot R_{2m})$, by the distributivity of concatenation over union, which is in the required form.
$R = R_1 + R_2$, then $R$ is already a sum of products.
$R = R1^{*}$ can be reduced to sum of products by repeated application of the law: $(R_1 + R_2)^* = (R_1^*\cdot R_2^*)^*$ and distributivity of concatenation over union.


Hence we can write each regular expression as sum of products. 

This is called the disjunctive normal form of regular expressions.