For $f\in L^2(\mathbb{C})$, $g=\begin{pmatrix} 1&\\z_0&1\end{pmatrix}$ acts on $f(z)$ as \begin{eqnarray*} (g.f)(z)=f(z-z_0). \end{eqnarray*} Let $B$ be an linear bounded operator defined on $L^2(\mathbb C)$ which commutes with the action of $g$. Then there exists $m(\xi)\in L^\infty(\mathbb C)$ such that \begin{eqnarray} \widehat{Bf}(\xi)=m(\xi)\widehat{f}(\xi), \end{eqnarray} where $\widehat{f}$ is the Fourier transform of $f$.
How to prove the above formula?
First, I assume that you mean that $B$ commutes with all operators of the form $f\mapsto g\cdot f$.
For any $a\in\mathbb C$, denote by $e_a$ the function $z\mapsto e^{a\cdot z}$, where $a\cdot z$ is the usual scalar product on $\mathbb C=\mathbb R^2$. By Plancherel, it is enough to show that f $T$ is an operator on $L^2(\mathbb C)$ which commutes with all multiplication operators $f\mapsto e_a f$, then $T$ is itself a multiplication operator by some $m\in L^\infty$.
Start with $f_0\in L^2$ such that the linear span of $\{ e_a f;\; a\in\mathbb C\}$ is dense in $L^2$; such an $f_0$ does indeed exist and satisfie $f_0(z)\neq 0$ almost everywhere (I take this as "known").
By assumption on $T$ we have $T(e_a f_0)=e_a Tf_0$ for all $a\in\mathbb C$. Since $f_0(z)\neq 0$ almost everywhere, it follows that $m:=\frac{Tf_0}{f_0}$ is well-defined almost everywhere and, for each fixed $a\in\mathbb C$: $$T(e_af_0)=m\times e_af_0\, . $$
Now, choose a countable dense set $D\subset \mathbb C$. Then the following holds true almost everywhere: $$\forall a\in D\;:\; T(e_af_0)=m\, e_af_0\, .$$ By linearity, it follows that if we set $V:={\rm span}\,(\{ e_af_0;\; a\in D\})$, then we have almost everywhere: $$\forall v\in V\;:\ T(v)=mv\, .$$ By assumption on $f_0$, the linear subspace $V$ is dense in $L^2$. Hence, for any $f\in L^2$, one can find a sequence $(v_n)\subset V$ such that $v_n\to f$ in $L^2$. Then $T(v_n)\to T(f)$ in $L^2$, and extracting subsequences if necessary we may assume that $v_n\to f$ almost everywhere and $T(v_n)\to T(f)$ almost everywhere. So we get $$\forall f\in L^2\;:\; T(f)=mf\quad\hbox{almost everywhere}$$ It remains to show that $m\in L^\infty$; but this is clear since multiplication by $m$ is now seen to be a bounded operator on $L^2$.