To solve the given functional equations, we need to find functions $f$ and $g$ that satisfy the following two conditions for all $x, y > 0$:

1.
$$(f(x) + y - 1)(g(y) + x - 1) = (x + y)^2$$

2.
$$(-f(x) + y)(g(y) + x) = (x + y + 1)(y - x - 1)$$

### Step 1: Analyze the First Equation

Consider the first equation:

$$(f(x) + y - 1)(g(y) + x - 1) = (x + y)^2$$

Assume $f(x) = x + c$ and $g(y) = y + d$ for some constants $c$ and $d$. Substitute these into the equation:

$$(x + c + y - 1)(y + d + x - 1) = (x + y)^2$$

This simplifies to:

$$(x + y + c - 1)(x + y + d - 1) = (x + y)^2$$

By comparing terms, we get:

$$x + y + c - 1 = x + y + d - 1 = x + y$$

Thus, $c = 1$ and $d = 1$.

### Step 2: Verify with the Second Equation

Now, substitute $f(x) = x + 1$ and $g(y) = y + 1$ into the second equation:

$$(-f(x) + y)(g(y) + x) = (x + y + 1)(y - x - 1)$$

Substituting $f$ and $g$, we have:

$$(-(x + 1) + y)((y + 1) + x) = (x + y + 1)(y - x - 1)$$

This further simplifies to:

$$(y - x - 1)(x + y + 1) = (x + y + 1)(y - x - 1)$$

Both sides of the equation are identical, confirming our solution satisfies the second equation.

### Conclusion

The functions $f(x) = x + 1$ and $g(y) = y + 1$ satisfy both functional equations. Therefore, the solution is:

$$\boxed{f(x) = x + 1 \text{ and } g(y) = y + 1}$$

These functions are defined and satisfy the given equations for all $x, y > 0$.