This is my first post, so I hope it is fitting for the forum and detailed enough.
Let $a,b\in\mathbb{C}$ where $b\neq0$. Consider the ring $R=\mathbb{C}[x,y]/I$, where $I=(x^2,xy,y^2,ax+by)$. I’m trying to show that $R\cong\mathbb{C}[t]/(t^2)$ and here is my strategy:
Define $\varphi:\mathbb{C}[x,y]\rightarrow\mathbb{C}[t]/(t^2)$ where $$f(x,y)\stackrel{\varphi}{\longmapsto} f(t,\frac{-a}{b}t)$$ If I can show $\ker\varphi=I$ and $\text{im}\varphi= \mathbb{C}[t]/(t^2) $ the assertion follows directly from the isomorphism theorem for rings.
For any $f\in \mathbb{C}[t]$ there are polynomials $q,r\in \mathbb{C}[t]$ such that $f(t)=t^2q(t)+r(t)$ with $\deg r\in\{0,1\}$ or $r\equiv 0$ And hence $f$ is represented by a constant- or first order polynomial in $\mathbb{C}[t]/(t^2)$. So we can quite easily hit any element $\alpha_0 t+\alpha_1\in\mathbb{C}[t]/(t^2)$ with $\alpha_0 x+\alpha_1\in\mathbb{C}[x,y]$. As such, $\text{im }\varphi=\mathbb{C}[t]/(t^2)$.
Note that $x^2$, $y^2$ and $xy$ all maps to $t^2=0$ in $\mathbb{C}[t]/(t^2)$ and $ab+bx\mapsto at+b(\frac{-a}{b}t)=0$ so $\ker\varphi\supseteq I$. The other inclusion however is where I'm in doubt. Here are my ideas that did not provide any realizations (at least for me):
(1) Generally, if $f=\sum_{i,j}\alpha_{i,j}x^{i}y^j\in\mathbb{C}[x,y]$, then $f(t,\frac{-a}{b}t)=\sum_{i,j}\alpha_{i,j}t^{i+j}=\alpha_{0,0}+(\alpha_{1,0}+\frac{-a}{b}\alpha_{0,1})t$ in $\mathbb{C}[t]/(t^2)$ - is this correct?
(2) If $f\in\ker\varphi$ then I need to consider $\alpha_{0,0}+(\alpha_{1,0}+\frac{-a}{b}\alpha_{0,1})t=p(t)t^2$.
(3) I am not completely convinced, but this seems to show that then $\alpha_{0,0}=0$. So if $(\alpha_{1,0}+\frac{-a}{b}\alpha_{0,1})=0$, then $f$ is a multiple of $x^2$, $y^2$ or $xy$. Otherwise, if $(\alpha_{1,0}+\frac{-a}{b}\alpha_{0,1})\neq0$, then $f$ is a multiple of $ax+by$
I feel like most of my argumentation is based on gut feeling, so I am far from convincing myself. I will appreciate any comments.
Everything seems fine until step $(3)$. In that step, you always have $(\alpha_{10}+\frac{-a}{b}\alpha_{01})=0$, hence you always have that $$ \alpha_{10}=\frac{a}{b}\alpha_{01}.$$ This implies that $f$ has the form $$ f(x,y)=\frac{a}{b}\alpha_{01}x+\alpha_{01}y +h(x,y) $$ with $h(x,y)\in (x^{2},y^{2},xy)$. Therefore $$ f(x,y)=\frac{\alpha_{01}}{b}(ax+by) +h(x,y) \in I. $$
Note that from $\alpha_{10}=\alpha_{01}=0$ (which I believe is what you meant in your first case) you cannot deduce that $f$ is a multiple of $x^{2}$, $y^{2}$ or $xy$, e.g. consider $f(x,y)=x^{2}+y^{2}$. And similarly from $\alpha_{10}\neq 0$ you cannot deduce that $f$ is a multiple of $ax+by$, e.g. $f(x,y)=ax+by+x^{2}$.