You can show that the functions $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \big( x + f ( y ) \big) = y + f ( x + 1 ) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ are exactly those of the form $ f ( x ) = A ( x ) + 1 $, where $ A $ is an additive involution, i.e. $ A ( x + y ) = A ( x ) + A ( y ) $ and $ A \big( A ( x ) \big) = x $ for all $ x , y \in \mathbb R $. If further regularity conditions like continuity, local boundedness, integrability or measurability are assumed for $ f $, then $ A $ will be regular, too. The only regular additive functions are those of the form $ A ( x ) = c x $ for some constant $ c \in \mathbb R $ (see here). Thus the only regular additive involutions are $ A ( x ) = x $ and $ A ( x ) = - x $, as we must have $ A \big( A ( 1 ) \big) = c ^ 2 = 1 $. Therefore, the only regular solutions to \eqref{0} are $ f ( x ) = x + 1 $ and $ f ( x ) = - x + 1 $. Without any further conditions on $ f $, one can use the axiom of choice to show that there are nonregular solutions, too (see Example 5.6 in this PDF file).
It's straightforward to check that if $ f $ is of the form $ f ( x ) = A ( x ) + 1 $ for some additive involution $ A $, then it satisfies \eqref{0}. We try to prove the converse.
Letting $ x = 0 $ in \eqref{0} we get
$$ f \big( f ( y ) \big) = y + f ( 1 ) \text . \tag 1 \label 1 $$
In particular, \eqref{1} shows that if $ f ( x ) = f ( y ) $ then $ x = y $. Letting $ y = 0 $ in \eqref{1} shows that $ f \big( f ( 0 ) \big) = f ( 1 ) $, and thus by injectivity, $ f ( 0 ) = 1 $. Hence, putting $ x = - 1 $ in \eqref{0} we have
$$ f \big( f ( y ) - 1 \big) = y + 1 \text . \tag 2 \label 2 $$
\eqref{2} shows that if we substitute $ x - 1 $ for $ x $ and $ f ( y ) - 1 $ for $ y $ in \eqref{0}, we get
$$ f ( x + y ) = f ( x ) + f ( y ) - 1 \text . \tag 3 \label 3 $$
Now, if we define $ A : \mathbb R \to \mathbb R $ with $ A ( x ) = f ( x ) - 1 $, then by \eqref{3} we can see that $ A $ is additive, and by \eqref{2} we can see that $ A $ is an involution. This proves what was desired.
As Marktmeister has mentioned in a comment, for the functional equation
$$ f \bigl ( f ( x ) - x \bigr ) = 2 x \tag 0 \label 0 $$
to make sense for a given $ f : \mathbb R ^ + _ 0 \to \mathbb R ^ + _ 0 $ and all $ x \in \mathbb R ^ + _ 0 $, one must have $ f ( x ) \ge x $ for all $ x \in \mathbb R ^ + _ 0 $. In particular, $ f \bigl ( f ( x ) - x \bigr ) \ge f ( x ) - x $ for all $ x \in \mathbb R ^ + _ 0 $, which by \eqref{0} gives $ f ( x ) \le 3 x $ for all $ x \in \mathbb R ^ + _ 0 $. Let's iterate this method: define the sequences $ ( a _ n ) _ { n = 0 } ^ \infty $ and $ ( b _ n ) _ { n = 0 } ^ \infty $ of positive real numbers recursively by $ a _ 0 = 1 $ and for any nonnegative integer $ n $, $ b _ n = \frac 2 { a _ n } + 1 $ and $ a _ { n + 1 } = \frac 2 { b _ n } + 1 $. Note that similar to what we did above, by \eqref{0} and mathematical induction, one gets $ a _ n x \le f ( x ) \le b _ n x $ for all $ x \in \mathbb R ^ + _ 0 $ and all positive integers $ n $. Therefore, if we prove that as $ n $ tends to infinity, $ a _ n $ increasingly goes to $ 2 $ and $ b _ n $ decreasingly goes to $ 2 $, then we end up with $ f ( x ) = 2 x $ for all $ x \in \mathbb R ^ + _ 0 $, which indeed gives a solution to \eqref{0}. For this purpose, first use mathematical induction to prove that for any nonnegative integer $ n $, $ 1 \le a _ n < 2 $ and $ 2 < b _ n \le 3 $. Then, note that for any nonnegative integer $ n $,
$$ 0 < 2 - a _ { n + 1 } = 2 - \left ( \frac 2 { \frac 2 { a _ n } + 1 } + 1 \right ) = \frac { 2 - a _ n } { 2 + a _ n } \le \frac { 2 - a _ n } { 2 + 1 } = \frac { 2 - a _ n } 3 $$
and
$$ 0 < b _ { n + 1 } - 2 = \left ( \frac 2 { \frac 2 { b _ n } + 1 } + 1 \right ) - 2 = \frac { b _ n - 2 } { b _ n + 2 } < \frac { b _ n - 2 } { 2 + 2 } = \frac { b _ n - 2 } 4 \text . $$
Hence, by mathematical induction, we get
$$ 0 < 2 - a _ n \le \frac { 2 - a _ 0 } { 3 ^ n } = \frac 1 { 3 ^ n } $$
and
$$ 0 < b _ n - 2 \le \frac { b _ 0 - 2 } { 4 ^ n } = \frac 1 { 4 ^ n } $$
for all nonnegative integers $ n $, which proves what was desired.
Concerning the part you added later, if $ f : \mathbb R ^ + _ 0 \to \mathbb R ^ + _ 0 $ satisfies
$$ f \bigl ( f ( x ) - \alpha x \bigr ) = \beta x \tag 1 \label 1 $$
for some constants $ \alpha , \beta \in \mathbb R ^ + $ and all $ x \in \mathbb R ^ + _ 0 $, the same method as above works. Note that for all $ x \in \mathbb R ^ + _ 0 $ and all $ \gamma \in \mathbb R ^ + $, $ f \bigl ( f ( x ) - \alpha x \bigr ) \ge \gamma \bigl ( f ( x ) - \alpha x \bigr ) $ implies $ f ( x ) \le \left ( \frac \beta \gamma + \alpha \right ) x $ (by \eqref{1}), and $ f \bigl ( f ( x ) - \alpha x \bigr ) \le \gamma \bigl ( f ( x ) - \alpha x \bigr ) $ implies $ f ( x ) \ge \left ( \frac \beta \gamma + \alpha \right ) x $. Therefore, similar to before, we can define $ ( a _ m ) _ { m = 0 } ^ \infty $ and $ ( b _ m ) _ { m = 0 } ^ \infty $ by $ a _ 0 = \alpha $, $ b _ m = \frac \beta { a _ m } + \alpha $ and $ a _ { m + 1 } = \frac \beta { b _ m } + \alpha $, and inductively show that $ a _ m x \le f ( x ) \le b _ m x $ for all $ x \in \mathbb R ^ + $ and all nonnegative integers $ m $. Taking limits, you can then find out that $ f ( x ) = c x $ gives the unique solution, where $ c $ is the positive root of $ c ^ 2 - \alpha c - \beta $, i.e. $ c = \frac { \alpha + \sqrt { \alpha ^ 2 + 4 \beta } } 2 $. I leave verifying the details to you. In the special case where $ \alpha = n $ and $ \beta = n + 1 $, we have $ c = n + 1 $.
Best Answer
For every $x$ you have two equations: $$\begin{cases} x f(x) + f(-x) = x \\ f(x) - x f(-x) = -x \end{cases}$$ where the second equation is obtained by plugging $-x$ in the place of $x$.
This gives you a linear system with two equations and two unknowns $f(x), f(-x)$. The solution is $$f(x)= \frac{x^2-x}{x^2+1}$$ Now, you can easily check that such a solution is well defined for all $x \in \Bbb R$ and satisfies the functional equation.