Given the binary operation x y = xy + x + y on the set of real numbers, find the inverse of x and solve x everse igoplus 3 = 19.

The identity element $e$ is such that for any $x \in R$, $x \oplus e = x$.

$$x \oplus e = xe + x + e$$

Setting this equal to $x$:

$$xe + x + e = x$$

Subtract $x$ from both sides:

$$xe + e = 0$$

Factor out $e$:

$$e(x+1) = 0$$

For this equation to hold for all $x \in R$ (where $x \neq -1$), the identity element must be $e=0$.
We can verify this: $x \oplus 0 = x(0) + x + 0 = 0 + x + 0 = x$.
So, the identity element is $e=0$.

The inverse $x^{-1}$ is such that $x \oplus x^{-1} = e$.
Since $e=0$:

$$x \oplus x^{-1} = 0$$

Using the definition of the operation:

$$x(x^{-1}) + x + x^{-1} = 0$$

Factor out $x^{-1}$ from the terms containing it:

$$x^{-1}(x+1) + x = 0$$

Subtract $x$ from both sides:

$$x^{-1}(x+1) = -x$$

Divide by $(x+1)$ to solve for $x^{-1}$ (provided $x+1 \neq 0$, i.e., $x \neq -1$):

$$x^{-1} = \frac{-x}{x+1}$$

The inverse of $x$ is $\boxed{\frac{-x}{x+1}}$ for $x \neq -1$.

The operation is $x \oplus y = xy + x + y$.
Substitute $y=3$ into the definition:

$$x \oplus 3 = x(3) + x + 3$$ $$x \oplus 3 = 3x + x + 3$$ $$x \oplus 3 = 4x + 3$$

We are given $x \oplus 3 = 19$:

$$4x + 3 = 19$$

Subtract 3 from both sides:

$$4x = 19 - 3$$ $$4x = 16$$

Divide by 4:

$$x = \frac{16}{4}$$ $$x = 4$$

The solution is $\boxed{x=4}$.