Step 1: Form the differential equation. Since the slope of the tangent at any point $(x, y)$ equals the sum of the coordinates, we have:

$$\frac{dy}{dx} = x + y$$

or

$$\frac{dy}{dx} - y = x$$

Step 2: Identify the type of differential equation. This is a linear differential equation of the form

$$\frac{dy}{dx} + Py = Q$$

where $P = -1$ and $Q = x$.

Step 3: Find the integrating factor.

$$IF = e^{\int P \, dx} = e^{\int -1 \, dx} = e^{-x}$$

Step 4: Apply the solution formula.

$$y \cdot IF = \int Q \cdot IF \, dx + C$$

which gives

$$y \cdot e^{-x} = \int x \cdot e^{-x} \, dx + C$$

Step 5: Integrate using integration by parts.

$$\int x \cdot e^{-x} \, dx = -(x+1)e^{-x}$$

Therefore,

$$y \cdot e^{-x} = -(x+1)e^{-x} + C$$

giving

$$y = Ce^x - x - 1$$

Step 6: Find $C$ using the condition that the curve passes through the origin $(0, 0)$. Substituting $x = 0$, $y = 0$:

$$0 = C \cdot e^0 - 0 - 1$$

so

$$C = 1$$

Step 7: Write the final equation. The equation of the curve is:

$$\boxed{y = e^x - x - 1}$$