If $a=0$ the system is trivial to solve, otherwise substituting $x\mapsto ax$, $y\mapsto ay$, $z\mapsto az$, $b_k \mapsto a^2 b_k$ and canceling out a factor of $a^2$ it can be assumed WLOG that $a=1$ and the system is:
$$
\begin{align}
x + yz &= b_1 \tag{1}\\
y + zx &= b_2 \tag{2}\\
z + xy &= b_3 \tag{3}
\end{align}
$$
This can be approached in several different ways, outlined below, but all of them eventually lead to a quintic equation which is not solvable by radicals in general, so the practical answer is to solve numerically.
Two variables can be eliminated between the three equations, leaving an equation in one variable alone. This was done step-by-step in @lonestudent's answer, deleted since but hopefully to be restored at some later time. Rather than duplicating the calculations here, this is the end result courtesy WA:
$$
x^5 - b_1 x^4 - 2 x^3 + (2 b_1 + b_2 b_3) x^2 + (-b_2^2 - b_3^2 + 1) x - b_1 + b_2 b_3 = 0 \tag{4}
$$
The symmetry of the system suggests combining the equations and expressing everything in terms of the elementary symmetric polynomials $e_1=x+y+z$, $e_2=xy+yz+zx$, $e_3=xyz$. After routine calculations, the following system in terms of $e_1,e_2,e_3$ is derived:
$$
\begin{align}
e_1 + e_2 &= u = b_1 + b_2 + b_3 \tag{5}
\\ e_1e_2 + e_1 e_3 + e_2 - 3e_3 &= v = b_1b_2 + b_2b_3 + b_3b_1 \tag{6}
\\ e_1^2e_3 - 2e_1e_3 + e_2^2 + e_3^2 - 2e_2e_3 + e_3 &= w = b_1b_2b_3 \tag{7}
\end{align}
$$
Eliminating (for example) $e_2, e_3$ between the equations gives an equation in $e_1$ alone, unsurprisingly also a quintic:
$$
e_1^5 - e_1^4 u + e_1^3 (-4 u + v - 6) + e_1^2 (4 u^2 + 14 u - v - w + 8) + e_1 (-8 u^2 - 4 u v - 12 u + 3 v + 6 w - 3) \\ + 4 u^2 + 4 u v + 3 u + v^2 - 3 v - 9 w = 0 \tag{8}
$$
If $x=0$ is a solution then it follows from $(2),(3)$ that $y=b_2, z=b_3$, and the system is compatible iff $(1)$ is satisfied i.e. $b_2b_3=b_1$. Same goes for $y,z=0$, which leaves the case $p=xyz \ne 0$ to consider. The equations can then be rewritten as:
$$
\begin{align}
x^2 - b_1 x + p &= 0 \tag{9}\\
y^2 - b_2 y + p &= 0 \tag{10}\\
z^2 - b_3 z + p &= 0 \tag{11}
\end{align}
$$
Eliminating $x,y,z$ between $(9),(10),(11)$ and $p = xyz$ gives a (laborious) sextic equation in $p$ alone, which reduces to a quintic after removing the ineligible root $p=0$.
Taking e.g. equation $(4)$, that's a quintic, which is not solvable by radicals in general. To show that $(4)$ is not some special form of quintic which happens to always be solvable, it is enough to exhibit a case where it is not solvable. For example, when $b_1 = 1$, $b_2 = 2$, $b_3 = 3$ the quintic is:
$$
x^5 - x^4 - 2 x^3 + 8 x^2 - 12 x + 5 = 0 \tag{12}
$$
It can be easily verified that the LHS in $(12)$ is negative at $x = -3, 1$ and positive at $x = 0, 2$, so the quintic has one negative and two positive real roots, and by Descartes' rule of signs these are all the real roots it can have, so the remaining two roots are complex conjugates.
It can also be verified that $(12)$ is irreducible, then it follows that it is not solvable by radicals because the full symmetric group $S_5$ is not solvable, and the Galois group of $(12)$ is $S_5$, based on:
Symmetric group of prime order
If $f$ is an irreducible polynomial of prime degree $p$ with rational coefficients and exactly two non-real roots, then the Galois group of $f$ is the full symmetric group $S_{p}$.
That the quintic $f(x)$ in $(4)$ is not solvable in general does not mean that it cannot be solvable in particular cases. A few examples where the quintic factors and is therefore solvable:
$b_2=b_3\text{:}\quad$ f(x) = $\,(x - 1)^2 (x^3 - (b_1 - 2) x^2 - (2 b_1 - 1) x + b_2^2 - b_1)\;$ ;
$b_2=b_1^2, b_3=b_1^3\text{:}\quad$ f(x) = $\,{(x - b_1) (x^4 - 2 x^2 + b_1^5 x - b_1^4 + 1)}\;$ ;
$b_1=1, b_2=0, b_3=-1\text{:}\quad$ f(x) = $\,{(x^2 - x - 1) (x^3 - x + 1)}\;$.
Algorithms exist to determine whether a given quintic is solvable (e.g. 1), but the calculations are extremely heavy, and if it turns out to be solvable it is even more difficult to determine the roots analytically, which makes it impractical in virtually all applications.
