Group of units of direct sum of rings is isomorphic to direct sum of the groups of units

Keep in mind that for a finite number of rings, the finite direct sum is isomorphic to the finite direct product. It is easier, then, to think of $R_1 \times R_2$ rather than of $R_1 \oplus R_2$. In particular, the elements of $R_1 \times R_2$ are pairs of the form $(r_1, r_2)$, the multiplication is $(r_1, r_2) (s_1, s_2) = (r_1 s_1, r_2 s_2)$ and the multiplicative neutral element is $(1,1)$.

It then follows easily that $(r_1, r_2)$ is invertible in $R_1 \times R_2$ if and only if there exist $(s_1, s_2) \in R_1 \times R_2$ such that $(r_1, r_2) (s_1, s_2) = (s_1, s_2) (r_1, r_2) = (1,1)$, which is equivalent to saying that $(r_1 s_1, r_2 s_2) = (s_1 r_1, s_2 r_2) = (1,1)$, which in turn is equivalent to saying that $r_1 s_1 = s_1 r_1 = 1 \in R_1$ and $r_2 s_2 = s_2 r_2 = 1 \in R_2$, which finally is equivalent to saying that $r_1$ is invertible in $R_1$ and $r_2$ is invertible in $R_2$.