You can find the solution below, and afterwards an explanation as of why your $u$ substitution did not work.
An alternative way is to write
$${\int_{0}^{2\pi}\frac{1}{1+\cos^2(x)} dx=4\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\cos^2(x)}dx}$$
$${=4\int_{0}^{\frac{\pi}{2}}\frac{\sin^2(x) + \cos^2(x)}{1+\cos^2(x)}dx}$$
Now, dividing the top and bottom by ${\cos^2(x)}$ gives
$${4\int_{0}^{\frac{\pi}{2}}\frac{\tan^2(x) + 1}{\sec^2(x) + 1} dx}$$
Now using some more trig identities, we get
$${=4\int_{0}^{\frac{\pi}{2}}\frac{\sec^2(x)}{\tan^2(x) + 2}dx}$$
$${=4\int_{0}^{\infty}\frac{1}{2+u^2}du}$$
Now, solving that integral
$${\int_{0}^{\infty}\frac{1}{2+u^2}du=\frac{1}{2}\int_{0}^{\infty}\frac{1}{1+\left(\frac{u}{\sqrt{2}}\right)^2}du}$$
Now do ${k=\frac{u}{\sqrt{2}}}$:
$${=\frac{1}{\sqrt{2}}\int_{0}^{\infty}\frac{1}{1+k^2}dk=\frac{\pi}{2\sqrt{2}}}$$
So putting the whole thing together:
$${=4\times \frac{\pi}{2\sqrt{2}}=\sqrt{2}\pi}$$
Which is the correct answer :)
EDIT: After some reconsideration, I don't believe the issue in your original ${u}$ substitution was really to do with injectivity at all. In fact injectivity is not a strict requirement for $u$ substitution. If we redo all the steps we just did, but don't change the domain of integration, you end up with
$${\int_{0}^{2\pi}\frac{1}{1+\cos^2(x)}dx=\int_{0}^{2\pi}\frac{\sec^2(x)}{\tan^2(x) + 2}dx}$$
You may be tempted to once again do ${u=\tan(x)}$ (which is essentially what you did I believe?) but one requirement for $u$ substitution that is very clearly needed is for $u$ to be continuous on the domain of integration (it needs to be differentiable, and so clearly continuity is a requirement!). ${u=\tan(x)}$ certainly is not continuous on ${(0,2\pi)}$, and so you really end up with an improper integral of sorts. Now, we could split up the integral instead, just as we would with any old improper integral:
$${\int_{0}^{2\pi}\frac{\sec^2(x)}{\tan^2(x) + 2}dx=\int_{0}^{\frac{\pi}{2}}\frac{\sec^2(x)}{\tan^2(x) + 2}dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\frac{\sec^2(x)}{\tan^2(x) + 2}dx + \int_{\frac{3\pi}{2}}^{2\pi}\frac{\sec^2(x)}{\tan^2(x) + 2}dx}$$
And now it is legitimate to do ${u=\tan(x)}$, since ${\tan(x)}$ will be differentiable and continuous on these domains (well, technically not at the endpoints - but you take a limit, as per the definition of how we deal with improper integrals). Indeed you will get the answer of ${\sqrt{2}\pi}$ if you evaluate this expression.
So if this was the issue, why did people jump to injectivity / bijectivity? Well there are some instances where this is (indirectly) the problem. An example:
$${\int_{0}^{2\pi}xdx}$$
Clearly the answer to this is ${2\pi^2}$. Now do the substitution ${u=\sin(x)}$ - our endpoints become ${\int_{0}^{0}}$... does this mean the integral is $0$? NO! Recall $u$ substitution only says that
$${\int_{a}^{b}f(\phi(x))\phi'(x)dx = \int_{\phi(a)}^{\phi(b)}f(u)du}$$
If you actually attempt to write ${\int_{0}^{2\pi}xdx}$ to match the form of the lefthand side so we can legit utilise $u$ substitution you will end up using some nasty ${\arcsin}$ rubbish - but the key point to take away is that the ${\arcsin}$ function only gives back principle values. ${\arcsin(\sin(x))}$ does not necessarily equal to ${x}$ for all ${x \in \mathbb{R}}$!. So in actuality you end up having a piecewise function to represent ${x}$, so you are forced to split the integral up. So in this case injectivity is in fact a sort of "requirement" indirectly (unless you split up the integral).
I hope this helped explain a bit better :)
Best Answer
To avoid confusion with limits, note that $$\int_0^{\pi} \dfrac{dx}{a+b \cos(x)} = \int_{\pi}^{2\pi} \dfrac{dx}{a+b \cos(x)}$$ Hence, we have $$I = \int_0^{2\pi} \dfrac{dx}{a+b \cos(x)} = 2\int_0^{\pi} \dfrac{dx}{a+b \cos(x)}$$ Now use your substitution $t = \tan(x/2)$ and note that $t$ goes from $0$ to $\infty$ as $x$ goes from $0$ to $\pi$.
EDIT
An easier way if you are familiar with a little bit of complex analysis is as follows.
Let $z = e^{ix}$. Then we have that $dz = iz dx$. Hence, \begin{align} I & = \int_0^{2 \pi} \dfrac{dx}{a+b \cos(x)} = \oint_{\vert z \vert = 1} \dfrac{dz}{iz \left( a + \dfrac{b}2 \left(z+ \dfrac1z \right)\right)}\\ & = \dfrac1i \oint_{\vert z \vert = 1} \dfrac{dz}{az + \dfrac{b}2 \left(z^2+1 \right)} = \dfrac2{ib} \oint_{\vert z \vert = 1}\dfrac{dz}{z^2 + \dfrac{2a}b z + 1}\\ & = \dfrac2{ib} \oint_{\vert z \vert = 1} \dfrac{dz}{\left(z + \dfrac{a}b \right)^2 + 1 - \dfrac{a^2}{b^2}} = \dfrac2{ib} \oint_{\vert z \vert = 1} \dfrac{dz}{\left(z + r + \sqrt{r^2-1} \right)\left(z + r - \sqrt{r^2-1} \right)} \end{align} where $r = \dfrac{a}b \in \mathbb{R}$.
First note that if $\vert r \vert \leq 1$, then $\left \vert r \pm \sqrt{r^2-1} \right \vert = 1$. Hence, the integral doesn't exist.
If $r > 1$, then $ \left \vert r + \sqrt{r^2 - 1} \right \vert > 1$ and $\left \vert r - \sqrt{r^2-1} \right \vert < 1$. Hence, by Cauchy integral formula, we have that $$I = \dfrac{2}{ib} 2 \pi i \dfrac1{2 \sqrt{r^2-1}} = \dfrac{2 \pi \, \text{sign}(a)}{\sqrt{a^2 - b^2}}$$
If $r < -1$, then $\left \vert r - \sqrt{r^2 - 1} \right \vert > 1$ and $\left \vert r + \sqrt{r^2-1} \right \vert < 1$. Hence, by Cauchy integral formula, we have that $$I = \dfrac{2}{ib} 2 \pi i \dfrac{-1}{2 \sqrt{r^2-1}} = \dfrac{2 \pi \, \text{sign}(a)}{\sqrt{a^2 - b^2}}$$ Hence, to summarize, $$I = \begin{cases} \dfrac{2 \pi \, \text{sign}(a)}{\sqrt{a^2 - b^2}} & \text{if } \vert a \vert > \vert b\\ \text{Does not exists} & \text{if } \vert a \vert \leq \vert b\end{cases}$$