For Question 1: First, there are idempotent ultrafilters, so some ultrafilters satisfy 1 in a very strong form. But 1 does not hold in general. The reason is that the semigroup $\beta\omega-\omega$ has subsemigroups $G$ that are groups of cardinality $2^{\mathfrak c}$ where $\mathfrak c$ is the cardinal of the continuum. (See Hindman and Strauss, "Algebra in the Stone-Cech Compactification", Theorems 2.7(d) and 2,25, and remember that all infinite closed subsets of $\beta\omega-\omega$ have cardinality $2^{\mathfrak c}$.) If the answer to Question 1 were always affirmative, then all elements of $G$ would be RK-below each other (and therefore isomorphic, but I don't need that). That's impossible as any ultrafilter is RK-above only $\mathfrak c$ others.

I expect the answer to Question 2 to be negative also, but I'll need to think some more to prove it. I also expect that there's an easier proof for Question 1. Finally, the answer to Question 1 becomes positive if either **a** or **b** is assumed to be a P-point.

Edit: A negative answer to Question 2 follows from the continuum hypothesis. Under CH (or certain weaker hypotheses, but not in ZFC alone), there exist P-point ultrafilters. Let **a** be one of these, and let **b** be an idempotent ultrafilter Rudin-Keisler above **a**. (It's provable in ZFC that any non-principal ultrafilter on $\omega$ is below an idempotent one.) Because **a** is a P-point, any sum $\mathbf a+\mathbf c$ (with **c** noprincipal) is isomorphic to the tensor product (also called Fubini product)
$$
\mathbf a\otimes\mathbf c=\{S\subseteq\omega\times\omega:\{x:\{y:(x,y)\in S\}\in\mathbf c\}\in\mathbf a\},
$$
and is therefore strictly above both **a** and **c** in the Rudin-Keisler order. In particular, $\mathbf a+\mathbf b\not\leq_{\text{RK}}\mathbf b$. But since **b** is idempotent, this gives $\mathbf a+\mathbf b\not\leq_{\text{RK}}\mathbf b+\mathbf b$ even though $\mathbf a\leq_{\text{RK}}\mathbf b$.

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