Answer
Yes,
$X=CB-A$
Work Step by Step
Multiply both sides with
$C$ from the left, and with
$B$ from the right
$ \begin{array}{rrlll}
\cdot C/ & C^{-1}(A+X)B^{-1} & = & I_{n} & /\cdot B\\
& C\cdot C^{-1}(A+X)B^{-1}\cdot B & = & C\cdot I_{n}\cdot B & \\
& I\cdot(A+X)\cdot I & = & CB & \\
& A+X & = & CB & /-X\\
& X & = & CB-A &
\end{array}$
Yes,
$X=CB-A$
