By using the Iwasawa decomposition, one obtains a (biinvariant) Haar measure on $G:=\mathrm{SL}(2,\mathbb{R})$ which can be symbolically written as $\mathrm{d}x=\mathrm{d}a\,\mathrm{d}n\,\mathrm{d}k$, the measures appearing on the righthand side being the usual ones. This actually means $$\int_G f(x)\,\mathrm{d}x=\int_K \int_N \int_A f(ank)\,\mathrm{d}a\,\mathrm{d}n\,\mathrm{d}k$$ for suitable $f$. (Here the order in $ank$, which is expressed by the notation $\mathrm{d}x=\mathrm{d}a\,\mathrm{d}n\,\mathrm{d}k$, is important, whereas the order of the triple integral is immaterial.) Analogously, one can define another Haar measure $\mathrm{d}_1 x=\mathrm{d}k\,\mathrm{d}n\,\mathrm{d}a$ so that $$\int_G f(x)\,\mathrm{d}_1 x=\int_A \int_N \int_K f(kna)\,\mathrm{d}k\,\mathrm{d}n\,\mathrm{d}a.$$ We must have $\mathrm{d}_1 x=c\cdot\mathrm{d}x$ for some $c>0$. Is $c$ equal to 1?

Yes, since you get one from the other by applying the inversion and there is a general rule for the inversion that says $\int_Gf(x^{1})dx=\int_Gf(x)dx\Delta(x)$, where $\Delta(x)$ is the modular function. 


Use Fubini to get: $I:=\int_G f(x) \mathrm{d}_1x = \int_A \int_N \int_K f(kna) \mathrm{d}k\mathrm{d}n\mathrm{d}a=\int_K \int_N \int_A f(kna) \mathrm{d}a\mathrm{d}n\mathrm{d}k$. Use unimodularity of K,N,A to get: $I=\int_K \int_N \int_A f(k^{1}n^{1}a^{1}) \mathrm{d}a\mathrm{d}n\mathrm{d}k=\int_K \int_N \int_A f((ank)^{1}) \mathrm{d}a\mathrm{d}n\mathrm{d}k$. Now use the unimodularity of $SL(2,\mathbb{R})$: $I=\int_G f(x^{1}) \mathrm{d}x=\int_G f(x) \mathrm{d}x$. “Using unimodularity” means using the identity given by anton: $\int_G f(x^{1}) \mathrm{d}x = \int_G f(x)\Delta(x) \mathrm{d}x$ where you use the left Haar measure (integrating $f(x^{1})$ with respect to the left Haar measure is the same as integrating $f$ with respect to the right Haar measure). As you noticed for these groups there exists a biinvariant Haar measure, thus the groups are unimodular (i. e. $\Delta(x)=1$). 

