Consider the function $$H(x):=\left.\begin{cases}0&\text{, if }x<0\\1&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }x\in\Omega:=(-1,1).$$
How can I show that there is no $g\in\mathcal L^2(\Omega)$ with $$\langle H,\varphi'\rangle_{L^2(\Omega)}=-\langle g,\varphi\rangle_{L^2(\Omega)}\tag1$$ for all $\varphi\in C_c^\infty(\Omega)$?
Clearly, if $g$ exists, then $$\varphi(0)=\langle g,\varphi\rangle\;\;\;\text{for all }\varphi\in C_c^\infty(\Omega)\tag2.$$ In particular, $$\varphi(0)=\langle g,\varphi\rangle\;\;\;\text{for all }\varphi\in C_c^\infty(\Omega\setminus\{0\})\tag3.$$
Now we know that $C_c^\infty(\Omega\setminus\{0\})$ is a dense subset of $L^2(\Omega\setminus\{0\})$ and hence $C_c^\infty(\Omega\setminus\{0\})^\perp=\overline{C_c^\infty(\Omega\setminus\{0\})}^\perp=\{0\}$. So, $$g(x)=0\;\;\;\text{for Lebesgue almost all }x\in\Omega\setminus\{0\}\tag4.$$
But I don't see that this is a contradiction to any of the assumptions.
(Supposing $g$ exists) note that since $\varphi$ is compactly supported, we actually have $$\varphi(0) = \langle g,\varphi\rangle$$ for all $\varphi$.
Now the main idea is to pick a sequence of functions $\varphi_n$ which breaks this by concentrating around the jump of $H$ at $x=0$.
For example, we can pick smooth functions such that for each $n$, $$\begin{gather*} \varphi_n(0)=1 \\ \varphi_n(x) \leq 1 \\ \text{supp}(\varphi) \subseteq [-1/n,1/n] \end{gather*}$$ You can imagine smoothing out an isosceles-triangle-shaped function with fixed height and decreasing width (in $n$), or use some standard bump function that you know. The important fact you should conclude is that $$\langle g,\varphi_n\rangle \xrightarrow{n\to\infty} 0.$$