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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 A₁,A₂,... have the exponential distribution and ε₁,ε₂,... take values 0,1 each with probability 1/2 (and all these rvs are independent), then X=∑_n2^1-nε_nA_n 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.

Edit 2: Apologies for keeping coming back to this one, but it seems interesting and my examples above are a special case of the following.

For any measurable subset A of the interval (0,1], you can define a random variable X_A with moment generating function E[exp(-λX_A)]=exp(-λ∫_Adx/(1+λx)). If you partition (0,1] into two measurable sets A,B and X_A,X_B are independent, then X_A+X_B has the exponential distribution. If A and B are unions of finitely many intervals then the moment generating functions will be rational functions of λ and its easy to make sure that X_A,X_B are not gamma distributed. My first example above is using the partition (0,1/2],(1/2,1]. The second one, in the edit, is partitioning (0,1] into the intervals (2^-n,2^1-n].

You can construct X_A as follows. Let T₁,T₂,… be independent with exponential distribution, and S_n=exp(-T₁-…-T_n). The number of S_n in a subset A of (0,1] will be Poisson with parameter ∫_Adx/x. If Y₁,Y₂,… are independent exponentially distributed then X_A=∑_n1_{Sn∈A}S_nY_n has the correct moment generating function. (I'll leave you work through the details...).

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 A₁,A₂,... have the exponential distribution and ε₁,ε₂,... take values 0,1 each with probability 1/2 (and all these rvs are independent), then X=∑_n2^1-nε_nA_n 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.

Edit 2: Apologies for keeping coming back to this one, but it seems interesting and my examples above are a special case of the following.

For any k>0 and measurable subset A of the interval (0,1], you can define a random variable X_A with moment generating function E[exp(-λX_A)]=exp(-λk∫_Adx/(1+λx)). If you partition (0,1] into two measurable sets A,B and X_A,X_B are independent, then X_A+X_B has the Gamma(k) distribution. If A and B are unions of finitely many intervals then the moment generating functions will be k^th powers of rational functions of λ and its easy to make sure that X_A,X_B are not gamma distributed. My first example above is using k=1 and the partition (0,1/2],(1/2,1]. The second one, in the edit, is partitioning (0,1] into the intervals (2^-n,2^1-n].

You can construct X_A as follows. Let T₁,T₂,… be independent with the Exp(k) distribution, and S_n=exp(-T₁-…-T_n). The number of S_n in a subset A of (0,1] will be Poisson with parameter ∫_Adx/x. If Y₁,Y₂,… are independent exponentially distributed then X_A=∑_n1_{Sn∈A}S_nY_n has the correct moment generating function. (I'll leave you work through the details...). Alternatively, the set {(S_n,Y_n):n≥1} is a Poisson point process with intensity ke^-y dy ds/s.

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 A₁,A₂,... have the exponential distribution and ε₁,ε₂,... take values 0,1 each with probability 1/2 (and all these rvs are independent), then X=∑_n2^1-nε_nA_n 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.

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 A₁,A₂,... have the exponential distribution and ε₁,ε₂,... take values 0,1 each with probability 1/2 (and all these rvs are independent), then X=∑_n2^1-nε_nA_n 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.

Edit 2: Apologies for keeping coming back to this one, but it seems interesting and my examples above are a special case of the following.

For any measurable subset A of the interval (0,1], you can define a random variable X_A with moment generating function E[exp(-λX_A)]=exp(-λ∫_Adx/(1+λx)). If you partition (0,1] into two measurable sets A,B and X_A,X_B are independent, then X_A+X_B has the exponential distribution. If A and B are unions of finitely many intervals then the moment generating functions will be rational functions of λ and its easy to make sure that X_A,X_B are not gamma distributed. My first example above is using the partition (0,1/2],(1/2,1]. The second one, in the edit, is partitioning (0,1] into the intervals (2^-n,2^1-n].

You can construct X_A as follows. Let T₁,T₂,… be independent with exponential distribution, and S_n=exp(-T₁-…-T_n). The number of S_n in a subset A of (0,1] will be Poisson with parameter ∫_Adx/x. If Y₁,Y₂,… are independent exponentially distributed then X_A=∑_n1_{Sn∈A}S_nY_n has the correct moment generating function. (I'll leave you work through the details...).

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.

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 A₁,A₂,... have the exponential distribution and ε₁,ε₂,... take values 0,1 each with probability 1/2 (and all these rvs are independent), then X=∑_n2^1-nε_nA_n 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.