Let $G$ be a Lie group. The adjoint representation $\text{Ad}:G \rightarrow \text{GL}(T_e G)$is given by $$ \text{Ad}(x):=T_e \mathcal{C}_x, $$ where $\mathcal{C}_x(y)=xyx^{-1}$.
Now suppose that $G=\text{SL}(n,\mathbb{R})$. Then I would think that $$ T_e \mathcal{C}_x (y) = \left.\frac{d}{dt}\right|_{t=0} x(e+t y)x^{-1} = \mathcal{C}_x(y). $$ Is this correct. If this is so, for which Lie groups can't I do this?
Yes, you are correct: if the group is realized as a subgroup $G \subset GL(n)$ then you can identify $T_eG \subset T_eGL(n) \equiv GL(n)$. Under such identification you will have $Ad(x)(y)=xyx^{-1}$