You can decompose the exponential distribution into a sum of two terms, which are not both gamma distributed.

Let A,B,ε be independent where A,B are exponentially distributed and ε takes the values 0,1 each with probability 1/2, and set X=A/2, Y=εB. You can calculate the moment generating functions of X and Y,
$$
E\left[\exp(-\lambda X)\right] = E\left[\exp(-(\lambda/2)A)\right]=1/(1+\lambda/2).
$$
$$
E\left[\exp(-\lambda Y)\right]=(1/2)E\left[\exp(-\lambda B)\right]+1/2=(2+\lambda)/(2+2\lambda).
$$
Then you can check the moment generating function function of X+Y, E[exp(-λ(X+Y)]=E[exp(-λX)]E[exp(-λY)]=1/(1+λ) to see that X+Y has the exponential distribution.

Edit:
After reading at Michael Lugo's response below, it might be more satisfying to have an answer where neither of X or Y are Gamma distributed. In fact, by iterating my argument above you can get the following example. If A<sub>1</sub>,A<sub>2</sub>,... have the exponential distribution and &epsilon;<sub>1</sub>,&epsilon;<sub>2</sub>,... take values 0,1 each with probability 1/2 (and all these rvs are independent), then X=&sum;<sub>n</sub>2<sup>1-n</sup>&epsilon;<sub>n</sub>A<sub>n</sub> has the exponential distribution (just check the moment generating function). By splitting this sum up into two smaller sums you can generate a whole load of counterexamples where neither term is gamma distributed.