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$\infty$ is not a number so $\arctan(\infty)$ isn't meaningful, nor does it have a reasonably standard interpretation. You can already tell something is wrong, because there's nothing in your argument that keeps it from applying for all values of $x$. One way to attempt to make an argument along the lines you sketch would be to perturb the problem slightly, then take the limit as the perturbation approaches $0$.

For example, $$\begin{align}
\arctan\left(\frac{x+\varepsilon}{x+1}\right) + \arctan\left(\frac{x+1}{x}\right)
& = \arctan\left(\frac{\frac{x+\varepsilon}{x+1} + \frac{x+1}{x}}{1 - \left(\frac{x+\varepsilon}{x+1}\right)\left(\frac{x+1}{x}\right)}\right) \\\\
& = \arctan\left(\frac{\frac{x+\varepsilon}{x+1} + \frac{x+1}{x}}{\frac{-\varepsilon}{x}}\right) \\\\
& = \arctan\left(-\frac{x^2}{\varepsilon(x+1)} - \frac{x}{x+1} - \frac{x+1}{\varepsilon}\right) \\\\
& = \arctan\left(-\frac{x^2+(x+1)^2}{\varepsilon(x+1)} - \frac{x}{x+1}\right) \\\\
\end{align}$$

Since the limit of $z^{-1}$ approaches $\infty$ when $z$ approaches $0$ from the right and $-\infty$ when approached from the left, the limit of the above expression is dependent on the sign of $\frac{x^2 + (x+1)^2}{x+1}$ and the direction we have $\varepsilon$ approach $0$. With this we know that the answer is at least $\pm\frac{\pi}{2}$.

Short of making the choices that give the right answer, it's not obvious how we should decide how $\varepsilon$ approaches $0$. This is the point where the fact that $$\arctan(A)+\arctan(B)=\arctan\left(\frac{A+B}{1-AB}\right)$$ is **false** starts being relevant. You can see this simply by setting $A=1$ and $B=2$. Because $\arctan$ is an odd function, the left hand side is necessarily positive, but the right hand side is $\arctan(-3)$ which is necessarily negative. The formula is only valid when $0 < 1 - AB$. For the above formulas to be valid, then, $\frac{-\varepsilon}{x}$ must be positive. For the given ranges, this means $\varepsilon$ must have the opposite sign as *both* $x$ and $x+1$ giving the desired positive $\frac{\pi}{2}$. We also see that for $x\in(-1,0)$, the constraint requires $\varepsilon$ to be positive but $x+1$ will positive as well which will lead to $-\frac{\pi}{2}$ which is indeed what you see in a plot.

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As an addendum, here's a geometric/trigonometric argument. First note that the domain constraints ensure that $x$ and $x+1$ always have the same sign, so the arguments to $\arctan$ are always positive. For $a$ and $b$ both positive, $\theta = \arctan(\frac{b}{a})$ is in the interval $[0,\frac{\pi}{2})$. We can view $a$ and $b$ as, respectively, the $x$ and $y$ coordinates of a point. $\theta$ is then the angle between the $x$-axis and the ray from the origin through the point $(a, b)$. If we then consider $\varphi = \arctan(\frac{a}{b})$, we can use the same logic to view it as the angle between the $x$-axis and the ray through $(b,a)$. But $(b,a)$ is the reflection of $(a,b)$ through the line $y=x$, so the angle between the $y$-axis (**note**, $y$-axis, not $x$-axis) and the ray through $(b,a)$ is $\theta$, therefore $\varphi = \frac{\pi}{2}-\theta$.

You can use vector algebra, complex exponentials, or trigonometric identities if you want a more symbolic rendition of this.