Attacking the problem backwards from Poisson equation to its radial symmetric equation is more intuitive in my opinion.
$\newcommand{\pp}[2]{\frac{\partial #1}{\partial #2}}$
$\newcommand{\pb}[2]{\frac{\partial }{\partial #2}\left(#1\right)}$
$\newcommand{\ppt}[2]{\frac{\partial^2 #1}{\partial #2^2}}$
Consider the Poisson's equation $\Delta v = v_{xx} + v_{yy} = g$ in polar coordinates $(r,\theta)$:
$$
r = \sqrt{x^2 + y^2} \text{ and } \theta = \arctan(y/x) + C
$$
We have:
$$\pp{r}{x} = \frac{x}{\sqrt{x^2+y^2}} = \frac{x}{r}, \text{ and } \pp{r}{y} = \frac{y}{r}
$$
$$\pp{\theta}{x} = \frac{-\frac{y}{x^2}}{1+\frac{y^2}{x^2}} = -\frac{y}{r^2}, \text{ and } \pp{\theta}{y} = \frac{x}{r^2}
$$
Now
$$
\pp{v}{x} = \pp{v}{r} \pp{r}{x} + \pp{v}{\theta} \pp{\theta}{x} = \pp{v}{r} \frac{x}{r} - \pp{v}{\theta} \frac{y}{r^2}
$$
and
$$
\begin{aligned}
v_{xx}= \pb{\pp{v}{x}}{x} &= \ppt{v}{r}\pp{r}{x}\frac{x}{r} + \pp{v}{r}\pb{\frac{x}{r} }{x} - \ppt{v}{\theta}\pp{\theta}{x}\frac{y}{r^2} - \pp{v}{\theta}\pb{\frac{y}{r^2}}{x}
\\
&= \ppt{v}{r}\frac{x^2}{r} + \pp{v}{r}\left(\frac{1}{r} - \frac{x^2}{r^3}\right)
+ \ppt{v}{\theta}\frac{y^2}{r^4} + \pp{v}{\theta}\frac{2xy}{r^4}
\end{aligned}\tag{1}
$$
Similarly we have:
$$
v_{yy} = \ppt{v}{r}\frac{y^2}{r} + \pp{v}{r}\left(\frac{1}{r} - \frac{y^2}{r^3}\right)
+ \ppt{v}{\theta}\frac{x^2}{r^4} - \pp{v}{\theta}\frac{2xy}{r^4}\tag{2}
$$
Exploiting $x^2+y^2 = r^2$, (1) plus (2) gives:
$$
\Delta v = \ppt{v}{r} + \frac{1}{r}\pp{v}{r}+\frac{1}{r^2}\ppt{v}{\theta}.
$$
If a function is radial symmetric, meaning it can be represented using only $r$ for it is a constant for fixed $r$, i.e., it is a constant on any fixed circle, and rotating does not change its value. Hence $$\displaystyle \ppt{v}{\theta} = 0.$$ And the Laplacian of a radial symmetric function is again radial symmetric for no involvement of $\theta$. Here I use $r$ for the $t$ in your question.
Let $v(x,y) =u(\sqrt{x^2 + y^2}) =u(r)$, $g(x,y) = f(r)$. Then
$$
u'' + \frac{1}{r}u' = f(r)
$$
Using your favorite method of solving ODE, here I cook up using product rule reverse direction:
$$
(ru')' = rf(r)
$$
Integrating:
$$ u'(r) = \frac{1}{r}(\int^r_0 sf(s) ds+ r u'(r)|_{r=0}) \tag{3}
$$
Since the limit exists when $r\to 0$:
$$
a = u'(0) = \lim_{r\to 0}\frac{\int^r_0 sf(s) ds}{r} = \lim_{r\to 0} rf(r) =0
$$
Now for $b$, since $v = 0$ when $r= 1$, hence $b = u(1) = 0$.
Hence $a = b =0$.
If you would like to get a form of the solution, integrating (3) again yields:
$$ u(r) - u(0) = \int^{r}_0 u'(\tau)d\tau = \int^{r}_0 \frac{1}{\tau}\int^{\tau}_0 sf(s) ds d\tau.
$$
