Let $I$ and $J$ be ideals of a ring $R=\mathbb{K}[X_1, \dots, X_m]/K$, quotient of a polynomial ring over a field $\mathbb{K}$.
Consider the map $$\begin{aligned} (R/I)^*\oplus (R/J)^*&\longrightarrow (R/(I+J))^*\\ (r, s)&\longmapsto (r\cdot s^{-1}) \end{aligned} $$ where $(R/I)^*$ are the units of the quotient ring.
Is this map surjective, in this generality?
If so, can you point out a reference, or say to me why this should be trivial?
If not, can you provide a counterexample?
If you had asked the question over any field and not specifically $\mathbb{K}$ (which I read as $\mathbb{R}$ or $\mathbb{C}$) I would have the following answer:
Take $R= \mathbb{Q}[X,Y]$, $I=(X^2-2)$, $J=(Y^2-3)$. Then in $R/(I+J)=\mathbb{Q}[\sqrt 2, \sqrt 3]$ we have $(Y-X)(Y+X)=Y^2-X^2=1$. It's impossible (in $R/(I+J)$) to write $Y-X$ as the product of two polynomials of which one is only in $X$ and the other only in $Y$, so it's not in the image.
For the case you posted I have the following similar idea:
Consider $R=\mathbb{C}[X,Y,Z,W]$, $I=(XY-1)$, $J=(ZW-1,XW+YZ)$.
Then in $R/(I+J)$ we have $(X+Z)\frac{Y+W}{2} = \frac{XY+XW+YZ+ZW}{2}=1$ but in neither $R/I$ nor $R/J$ the element $X+Z$ can be a unit.
Until someone can make it clear why $X+Z$ can't factor into elements which are units in the respective rings this answer is, however, only partial.