Determining if a symmetric matrix is positive definite

Yes. Your matrix can be written as $$(a+b)I + a ee^T$$ where $I$ is the identity matrix and $e$ is the vector of ones. This is a sum of a symmetric positive definite (SPD) matrix and a symmetric positive semidefinite matrix. Hence it is SPD.

In addition to Carl's answer (+1) you can use the Gershgorin circle theorem which is one I'll often try when I see diagonally dominant matrices like this one.

For each row, the sum of the absolute values of the off-diagonal entries is $2a$. The theorem then says that every eigenvalue is in a closed disc of radius $2a$ centered at the diagonal elements $2a + b$, so for any eigenvalue $\lambda$ we have $\lambda \geq b > 0$.

$$\begin{bmatrix} 2a+b & a & a\\ a & 2a+b & a\\ a & a & 2a+b\end{bmatrix} = (a+b) \, \mathrm I_3 + a \, 1_3 1_3^\top = (a+b) \, \mathrm I_3 + 3a \left(\frac{\, 1_3}{\sqrt{3}}\right) \left(\frac{\, 1_3}{\sqrt{3}}\right)^\top$$

Hence, the eigenvalues are $4a + b$ and $a+b$, with multiplicities $1$ and $2$, respectively. Since $a, b > 0$, both eigenvalues are positive, i.e., the given symmetric matrix is positive definite.