Indeed, there is at most one root on $x\in[0,\infty)$ -- actually, for each real $a$. Consider the equivalent equation \begin{equation} r(x):=\frac{f(x)}{g(x)}=\frac{\sqrt\pi}b \tag{1} \end{equation} for real $x>0\vee(-a)$, where $0<b<1$ and \begin{equation} f(x):=\frac{e^{-x^2}}{x+a},\quad g(x):=1-\text{erf}(x). \end{equation} Consider the "derivative ratio" \begin{equation} \rho(x):=\frac{f'(x)}{g'(x)}=\frac{\sqrt{\pi } \left(1+2 a x+2 x^2\right)}{2 (x+a)^2}. \end{equation} Then \begin{equation} \rho'(x)=\frac{\sqrt{\pi } \left(a^2+a x-1\right)}{(x+a)^3}, \end{equation} so that, for real $x>0\vee(-a)$, we have $$\rho'(x)>0 \iff a>1\text{ or } \big(0 < a \le 1\ \&\ x > x_a:=(1 - a^2)/a\big).$$
Now we are ready to use l'Hospital-type rules for monotonicity.
Case 1: $a\le0$. Then $\rho$ is (strictly) decreasing (on the entire interval $(0\vee(-a),\infty)$), whence, by Proposition 4.1 in the mentioned paper, $r$ is decreasing, and so, equation (1) has at most one root (in $(0\vee(-a),\infty)$). In fact, in this case there is exactly one root, since $r(0+)=\infty$ and $r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$.
Case 2: $0<a\le1$. Then $\rho$ is decreasing on $(0\vee(-a),x_a)$ and increasing on $(x_a,\infty)$), whence, by Proposition 4.3 in the same paper, $r$ is decreasing on $(0,c_a)$ and increasing on $(c_a,\infty)$, for some real $c_a\ge0$. So, $r<r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$ on $[c_a,\infty)$. Hence, equation (1) has at most one root in $(0\vee(-a),\infty)$, and that possible root must be in $(0,c_a)$. In fact, in this case as well there is exactly one root, since $r(0+)=\infty>\frac{\sqrt\pi}b$ and $r(c_a)<r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$.
Case 3: $a>1$. Then $\rho$ is increasing (on the entire interval $(0\vee(-a),\infty)$), whence, by the mentioned Proposition 4.1, $r$ is increasing. So, $r<r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$. Thus, equation (1) has no roots (in $(0\vee(-a),\infty)$) in this case.