I would like to understand in some detail why this Runge-Kutta method is explicit if and only if $c_1=0$. Why in this case the r.h.s. doesn't contain $y_{n+1}$ ?
EDIT $c_1=0$
EDIT 2
I would also like to know if this implicit Euler method
$$\begin{array}{c|c}1&1\\\hline&1\end{array}$$ is implicit in $k_1$ or also in $y_{n+1}$.
In other words, is it possible to obtain all these 4 possibilities?
Like being $(i)$ implicit for $k_1$ and explicit for $y_1$,
$(ii)$ implicit for $k_1$ and implicit for $y_1$,
$(iii)$ explicit for $k_1$ and explicit for $y_1$,
$(iv)$ explicit for $k_1$ and implicit for $y_1$
all of this for some suitable $s\geq 1$ stage method or perhaps with other index $i$ than $i=1$ ?
In other words do we/can we distinguish already in the definition in which variable namely whether $k_i$ or $y_i$ is the method implicit and in which it is explicit ?
Still, in your formula in the comment $y_{n+1}=y_n+hk_1$ the r.h.s. doesn't contain $y_{n+1}$.
In explicit one-step or Runge-Kutta methods, at the start of each step none of the slopes inside the step are known. Thus the first stage is always $k_1=f(x_n,y_n)$, that is, $c_1=0$. There is no information to reach inside the step interval.
The Euler methods are at an intersection of ideas that become distinct for higher-order methods, a conceptional twilight, undifferentiated embryonic forms. They are Runge-Kutta, but also linear multi-step. Their direct formulation à la multi-step is more intuitive than the separation of stages and composition that is usually done in the Runge-Kutta methods.
In general, when the explicit or implicit computation of $k_1, ..., k_s$ is finished, then the composition $y_{n+1}=y_n+h(b_1k_1+...+b_sk_s)$ is an explicit formula.
However, as already mentioned, in the one-stage methods $\begin{array}{c|c}\theta&\theta\\\hline &1\end{array}$ the composition is a direct linear relation between $k_1$ and $y_{n+1}$, one can be eliminated against the other. In this sense if $k_1$ is implicit, then so is $y_{n+1}$ and v.v. and inverse.