Let $Y_1 = aX_1 \sim \text{Bernoulli}(p)$ and $Y_2 = bX_2 \sim \text{Bernoulli}(p)$, what is the PMF of $Z = Y_1 + Y_2$?

My attempt:

$$p_{Y_1}(x) = \begin{cases}p & \text{if } x = a\\1 - p & \text{if } x = 0\end{cases}$$

$$p_{Y_2}(x) = \begin{cases}p & \text{if } x = b\\1 - p & \text{if } x = 0\end{cases}$$

Since for $X_1 + X_2$ the support of the Binomial r.v. is $\{0,1,2\}$, $Z$ should be defined on $\{a, b, a+b\}$(?). The convolution of $X_1 + X_2$ is $\begin{bmatrix}p^2 & 2p(1-p) & (1 - p)^2\end{bmatrix}$. Then

$$p_{Z}(x) = \begin{cases}p^2 & \text{if } x = a\\2p(1-p) & \text{if } x = a+b\\(1-p)^2 & \text{if } x = b\\\end{cases}$$

I test my result by comparing it to the expected value

$$E[Z] = aE[X_1] + bE[X_2] = (a + b)p$$

\begin{align*}E[Z] &= \sum_{x} p_Z(x)x\\
&= p_Z(a)a + p_Z(a+b)(a+b) + p_Z(b)b\\
&= p^2a + 2p(1-p)(a+b) + (1-p)^2b\\
&\neq (a+b)p
\end{align*}

Since a direct computation of the expected value gives a different result, my PMF is wrong.