Möbius tranformation taking reals to reals can be written with real coefficients I'm working on this one (from Ahlfors' Complex Analysis):

A Fractional Linear Transformation of form $\displaystyle T(z) = \frac{a z + b}{c z + d}$ which takes the real numbers into the real numbers can be written in a way where all the coefficients are real.

I'm pretty sure I know a way to get the answer.

First, the thing is a constant exactly when $ad - bc = 0$. Then you're done. If not, some coefficient is not zero. in that case, you can change variables and stuff around in a way so that you can assume WLOG $a \ne 0$.

Then, you pick three different real-valued inputs and show that all the three remaining coefficients $b, c, d$ are real-valued.

My question is:

Is there an easier way to do this, without using more advanced complex analysis (which I don't know yet)
It turns out that this is not an answer to the OP's question. I will leave this answer nevertheless, perhaps someone might find it useful for something. This was my fault for not reading the question more carefully

I will start the answer with an exercise. We will assume throughout that $a\neq 0$, and $ad-bc\neq 0$. Throughout, $L$ will denote a linear fractional transformation that sends extended reals to the extended reals, that satisfies the conditions above.

Exercise: Let $(z_1,z_2,z_3)$, and $(w_1,w_2,w_3)$ be two triples of extended complex numbers (the complex numbers with the point at infinity include). Then their is a unique linear fractional transformation, $L$ such that $L(z_i)=w_i,i=1,2,3.$ 

Now to take on the literal statement of the question, if a linear fractional transformation takes the reals to the reals, then it is a linear function (and takes infinity to infinity), but this is not so interesting. More interesting (the one I assume that you want to prove) is that a LFT that takes the extended reals to the extended reals (the reals with infinity included) mat be written to have all real coefficients. Since by your remark, that we may assume $a\neq 0$, multiply the numerator and denominator by $\frac{1}{a}$, so now without loss of generality, we may assume that $a=1$.

We will now invoke our exercise. If we say where we send $\{0,1,\infty\}$, we completely determine the LFT. Again, if we send infinity to infinity, we have a linear function. Let us assume that we send infinity to some real number. If $L(z)=\frac{z+b}{cz+d}$, then $L(\infty)=\frac{1}{c}$. If we assume that $L$ sends extended reals to extended reals, the, $\frac{1}{c}$ is real and therefore ,so is $c$. Likewise $L(0)=\frac{b}{d}$, which is also real if $d\neq 0$(we will also need to deal with the case when $d=0$. 
Since our linear fractional transformation does not send infinity to infinity, the number, $-\frac{d}{c}$ is real, since $L(-\frac{d}{c})=\infty$. Since $c$ is real, so is $d$. Since $\frac{b}{d}$ is real, so is $b$, since $d$ is real. Now suppose that $d=0$. Then the LFT is of the form, $\frac{z+b}{cz}=\frac{1}{c}+\frac{b}{cz}$. We still know that $c$ is real, the previous argument for this did not require $d\neq 0$. Now pick $z=1$ and plug it into $\frac{1}{c}+\frac{b}{cz}$, which concludes this argument.
It turns out that this is not an answer to the OP's question. I will leave this answer nevertheless, perhaps someone might find it useful for something. This was my fault for not reading the question more carefully

I will start the answer with an exercise. We will assume throughout that $a\neq 0$, and $ad-bc\neq 0$. Throughout, $L$ will denote a linear fractional transformation that sends extended reals to the extended reals, that satisfies the conditions above.

Exercise: Let $(z_1,z_2,z_3)$, and $(w_1,w_2,w_3)$ be two triples of extended complex numbers (the complex numbers with the point at infinity include). Then their is a unique linear fractional transformation, $L$ such that $L(z_i)=w_i,i=1,2,3.$ 

Now to take on the literal statement of the question, if a linear fractional transformation takes the reals to the reals, then it is a linear function (and takes infinity to infinity), but this is not so interesting. More interesting (the one I assume that you want to prove) is that a LFT that takes the extended reals to the extended reals (the reals with infinity included) mat be written to have all real coefficients. Since by your remark, that we may assume $a\neq 0$, multiply the numerator and denominator by $\frac{1}{a}$, so now without loss of generality, we may assume that $a=1$.

We will now invoke our exercise. If we say where we send $\{0,1,\infty\}$, we completely determine the LFT. Again, if we send infinity to infinity, we have a linear function. Let us assume that we send infinity to some real number. If $L(z)=\frac{z+b}{cz+d}$, then $L(\infty)=\frac{1}{c}$. If we assume that $L$ sends extended reals to extended reals, the, $\frac{1}{c}$ is real and therefore ,so is $c$. Likewise $L(0)=\frac{b}{d}$, which is also real if $d\neq 0$(we will also need to deal with the case when $d=0$. 
Since our linear fractional transformation does not send infinity to infinity, the number, $-\frac{d}{c}$ is real, since $L(-\frac{d}{c})=\infty$. Since $c$ is real, so is $d$. Since $\frac{b}{d}$ is real, so is $b$, since $d$ is real. Now suppose that $d=0$. Then the LFT is of the form, $\frac{z+b}{cz}=\frac{1}{c}+\frac{b}{cz}$. We still know that $c$ is real, the previous argument for this did not require $d\neq 0$. Now pick $z=1$ and plug it into $\frac{1}{c}+\frac{b}{cz}$, which concludes this argument.