To solve the functional equation

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

for all real numbers $x$ and $y$, we need to find all possible functions $f : \mathbb{R} \rightarrow \mathbb{R}$ that satisfy this condition.

Step 1: Initial Evaluation

Let's substitute $y = 0$ into the equation:

$$f(xf(x)) = f(x^2) + f(0) + x - 1$$

We'll refer to this as Equation (1).

Step 2: Evaluate for Specific Values

Substitute $x = 0$ into the given equation:

$$f(2y) = f(0) + f(y) + y - 1$$

We'll refer to this as Equation (2).

Set $y = 0$ in Equation (2):

$$f(0) = 2f(0) - 1 \implies f(0) = 1$$

Step 3: Simplify Using Known Values

Now that we know $f(0) = 1$, rewrite Equation (2) as:

$$f(2y) = f(y) + y$$

Call this Equation (3).

Step 4: Further Substitution

Next, use $x = 1$ in our initial equation:

$$f(f(1) + 2y) = f(1) + f(y) + 1 + y - 1$$

This simplifies to:

$$f(f(1) + 2y) = f(1) + f(y) + y$$

From Equation (3) $f(2y) = f(y) + y$, try $f(x) = x + 1$.

Verification

Assume $f(x) = x + 1$ and check it satisfies the original equation:

Evaluate $f(xf(x) + 2y) = f(x(x + 1) + 2y) = f(x^2 + x + 2y)$. Since $f(x) = x + 1$, this becomes:

$$x^2 + x + 2y + 1$$

Now, evaluate the right side:

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

Both sides are equal, therefore $f(x) = x + 1$ is indeed a solution.

Conclusion

The solution to the functional equation is:

$$\boxed{f(x) = x + 1}$$