My previous answer (below) only showed that an ideal gas can undergo an irreversible process that is both isothermal and adiabatic. However OP's question seems to be about the feasibility of two distinct processes, one isothermal but not adiabatic and another adiabatic but not isothermal, between the same initial and final states. This is possible as shown in T-S diagram below. 
----------------Previous answer---------------------
In an irreversible process it is possible for two distinct states to be connected by isothermal and adiabatic processes. Let the two states have pressure and volume $(p_1,V_1)$ and $(p_2,V_2)$ respectively. The for an isothermal process:
\begin{align}
p_1V_1=p_2V_2=\alpha
\end{align}
where $\alpha$ is a constant that depends on the magnitude of the temperature being held constant. For a reversible adiabatic process, which is also an isentropic process, we have:
\begin{align}
p_1V_1^\gamma=p_2V_2^\gamma=\beta
\end{align}
where $\beta$ is a constant which depends on the magnitude of the entropy being held constant. However an irreversible adiabatic process is not isentropic, and for this case:
\begin{align}
p_1V_1^\gamma & =\beta_1 \\
p_2V_2^\gamma & =\beta_2
\end{align}
where $\beta_1\neq \beta_2$, because entropy of the two states is different. Thus we have:
\begin{align}
\frac{V_2}{V_1} & =\left( \frac{\beta_2}{\beta_1} \right)^{1/(\gamma-1)} \\
\frac{p_2}{p_1} & =\left( \frac{\beta_2}{\beta_1} \right)^{1/(1-\gamma)}
\end{align}
Therefore for the set of initial and final states satisfying above relation, an isothermal and (irreversible) adiabatic curve can pass through the same set of initial and final states. This presumes that the same system can be run in reversible and irreversible modes between a given set of initial and final states.
As regards work done, Chester Miller has already answered your question.
